Chemistry 01 some basic concept of chemistry

some basic concept of chemistry

1.Calculate the molecular mass of the following:
(i)HO2
(ii)CO2
(iii)CH4

(i) HO2
The molecular mass of water, HO2
= (2 × Atomic mass of hydrogen) + (1 × Atomic mass of oxygen)
= [2(1.0084) + 1(16.00 u)]
= 2.016 u + 16.00 u
= 18.016
= 18.02 u

(ii)CO2
The molecular mass of carbon dioxide, CO2
= (1 × Atomic mass of carbon) + (2 × Atomic mass of oxygen)
= [1(12.011 u) + 2 (16.00 u)]
= 12.011 u + 32.00 u
= 44.01 u

(iii) CH4
The molecular mass of methane, CH4
= (1 ×Atomic mass of carbon) + (4 × Atomic mass of hydrogen)
= [1(12.011 u) + 4 (1.008 u)]
= 12.011 u + 4.032 u
= 16.043 u

2.Calculate the mass percent of different elements present in sodium sulphate ( Na 2 SO 4)

The molecular formula of sodium sulphate is Na2 SO4
Molar mass of Na2 SO4 =[(2 ×23.0) + (32.066) + 4 (16.00)]
142.066 g
Mass percent of an element =( Massofthat elementin thecompound /Molar massof thecompound ) × 100

∴ Mass percent of sodium:

= (46.0 g /142.066g ) 100

=32.379

=32.4%

Mass percent of sulphur:

=(32.066g/142.066g ) 100

=22.57

=22.6 %

Mass percent of oxygen:

=(64.0g/142.066g )100

=45.049

=45.05%

3.Determine the empirical formula of an oxide of iron which has 69.9% iron and 30.1% dioxygen by mass.

% of iron by mass 69.9 % [Given]
% of oxygen by mass 30.1 % [Given]
Relative moles of iron in iron oxide:
=  %of iron by mass / Atomicmassof iron
= 69.9/55.85
=1.25

Relative moles of oxygen in iron oxide:
= %of oxygen by mass / Atomic massor oxygen
= 30.1 / 16.00
= 1.88

Simplest molar ratio of iron to oxygen:
= 1.25 : 1.88
=1:1.5
= 2 : 3
∴ The empirical formula of the iron oxide is   Fe2O3 .

4.Calculate the amount of carbon dioxide that could be produced when
(i) 1 mole of carbon is burnt in air.
(ii) 1 mole of carbon is burnt in 16 g of dioxygen.
(iii) 2 moles of carbon are burnt in 16 g of dioxygen.

The balanced reaction of combustion of carbon can be written as:
C+ O2 → CO2
(i) As per the balanced equation, 1 mole of carbon burns in 1 mole of dioxygen (air) to produce 1 mole of carbon dioxide.

(ii) According to the question, only 16 g dioxygen is available. Hence, it will react with 0.5 mole of carbon to give 22 g of carbon dioxide. Hence, it is a limiting reactant.

(iii) According to the question, only 16 g dioxygen is available. It is a limiting reactant. Thus, 16 g of dioxygen can combine with only 0.5 mole of carbon to give 22 g of carbon dioxide.

5.Calculate the mass of sodium acetate (CH3COONa) required to make 500 mL of 0.375 molar aqueous solution. Molar mass of sodium acetate is 82.0245 g mol-1

0.375 M aqueous solution of sodium acetate
=1000 mL of solution containing 0.375 moles of sodium acetate
∴Number of moles of sodium acetate in 500 mL
=(0.375 /1000 ) 500
= 0.1875 mole
Molar mass of sodium acetate = 82.0.245 g mole-1 (Given)
∴Required mass of sodium acetate = (82.0245 g mol-1) (0.1875 mole)
= 15.38 g

6.Calculate the concentration of nitric acid in moles per litre in a sample which has a density, 1.41g mL-1 and the mass percent of nitric acid in it being 69%

Mass percent of nitric acid in the sample = 69% [Given]
Thus, 100 g of nitric acid contains 69 g of nitric acid by mass.
Molar mass of nitric acid (HNO3)

= {1 + 14 + 3 (16)} g mol-1
= 1 + 14 + 18
= 63 g mol-1
∴ Number of moles in 69 g of HNO3
69g/63g mol -1
=1.095mol

Volume of 100g of nitric acid solution
=Massof solution /density of solution
=100g/1.41gmL-1

=70.92mL = 70.92 × 10 -3 L

Concentration of nitric acid

= 1.095 mole /70.92 × 10 -3 L

=15.44mol /L
∴Concentration of nitric acid = 15.44 mol/L

7.How much copper can be obtained from 100 g of copper sulphate (CuSO4)?

1 mole of CuSO4 contains 1 mole of copper

Molar mass of CuSO4= (63.5) + (32.00) + 4(16.00)

= 63.5 + 32.00 + 64.00

= 159.5 g
159.5 g of CuSO4contains 63.5 g of copper

⇒100 g of CuSO4will contain 63.5 100g / 159.5 of copper.

∴ Amount of copper that can be obtained from 100g CuSO4 = (63.5 × 100) /159.5

=39.81 g

8.Determine the molecular formula of an oxide of iron in which the mass percent of iron and oxygen are 69.9 and 30.1 respectively. Given that the molar mass of the oxide is 159.69g mol-1 

Mass percent of iron (Fe) = 69.9% (Given)
Mass percent of oxygen (O) = 30.1% (Given)

Number of moles of iron present in the oxide =69.90 /55.85 =1.25

Number of moles of oxygen present in the oxide =30.1 /16.0 =1.88

Ratio of iron to oxygen in the oxide,
= 1.25 : 1.88

=1.25 /1.25   :  1.88/1.25

=1:1.5

=2:3

∴The empirical formula of the oxide is Fe2O3

Empirical formula mass of Fe2O3  = [2(55.85) + 3(16.00)] g

Molar mass of Fe2O3= 159.69 g

∴n =molar mass /Empiricalformula mass  =159.69g/159.7 g =0.999

=1(approx) 

Molecular formula of a compound is obtained by multiplying the empirical formula with n. Thus, the empirical formula of the given oxide is Fe2O3 and n is 1.
Hence, the molecular formula of the oxide is Fe2O3.

9.Calculate the atomic mass (average) of chlorine using the following data:

               % Natural Abundance          Molar Mass
                    35CI             75.77            34.9689
                    37CI             24.23            36.9659

The average atomic mass of chlorine

= [ (The average atomic mass of chlorine of 35 CI ) (molar mass of  35 CI  ) + (Fractionalabundance of 37 CI ) (Molar mass of 37 CI ) ]

=[ { ( 75.77/100) (34.9689 u) }  + { (24.23/100)(36.9659)} ] 

=26.4959 + 8.9568

=35.4527 u

10.In three moles of ethane (C2H6), calculate the following:
(i) Number of moles of carbon atoms.
(ii) Number of moles of hydrogen atoms.
(iii)Number of molecules of ethane.

(i) 1 mole of C2H6 contains 2 moles of carbon atoms.
∴Number of moles of carbon atoms in 3 moles of C2H6.
= 2 × 3 = 6
(ii) 1 mole of C2Hcontains 6 moles of hydrogen atoms.
∴Number of moles of carbon atoms in 3 moles of C2H6.
= 3 × 6 = 18
(iii) 1 mole of C2H6contains 6.023 ×1023 molecules of ethane.
×Number of molecules in 3 moles of C2H6
= 3 × 6.023 × 1023 = 18.069 ×1023

11.What is the concentration of sugar (C12H22O11) in mol L-1 if its 20 g are dissolved in enough water to make a final volume up to 2 L?

Molarity (M) of a solution is given by,

= Number of molesof solute / Volumeof solution in Litres

= Massof sugar/molar massof sugar / 2L

=20g / [ (12× 12)+ ( 1 ×22 )+(11 × 16) ] g / 2L

=(20g / 342g  ) / 2L

=0.0585 mol / 2L

=0.02925mol L -1

∴Molar concentration of sugar = 0.02925mol L -1

12.If the density of methanol is 0.793 kg L -1 , what is its volume needed for making 2.5 L of its 0.25 M solution?

Molar mass of methanol (CH3OH) = (1 × 12) + (4 ×1) + (1 × 16) 

= 32 g mol -1

=0,032 kg mol -1

Molarity of methanol solution = 0.793 kg L -1 / 0.032 hg mol -1

= 24.78 mol L -1

Since density is mass per unit volume)
Applying,
M1V1 = M2V2

(Given solution) (Solution to be prepared)

(24.78 mol L-1 ) V1 = (2.5 L) (0.25 mol L-1 )

V1 = 0.0252 L

V1 = 25.22 mL

13.Pressure is determined as force per unit area of surface. The SI unit of pressure, Pascal is as shown below:

1Pa = 1N m-2
If mass of air at sea level is 1034 g cm-2 , calculate the pressure in Pascal.

Pressure is defined as force acting per unit area of the surface

P =F/A

= (1034 g  × 9.8 ms -2 / cm2  ) × (1 kg /1000g  ) × (100)2 cm2 / 1m2

= 1.01332  × 10  kg m -1 s -2

We know,

1N =1kg ms -2

Then,

 1pa = 1Nm -2 = 1 kg m -2 s -2

1pa = 1 kg m -1 s -2

∴ Pressure  = 1.01332 × 105  pa

14.What is the SI unit of mass? How is it defined? 

The SI unit of mass is kilogram (kg). 1 Kilogram is defined as the mass equal to the mass of  the international prototype of kilogram.

15.Match the following prefixes with their multiples

 PrefixesMultiples
  i) micro106
  ii)deca109
  iii)mega10-6
  iv)giga10-15
  v)femto 10
 PrefixesMultiples
  i) micro10-6
  ii)deca10
  iii)mega106
  iv)giga109
  v)femto 10-15

16.What do you mean by significant figures?

Significant figures are those meaningful digits that are known with certainty. They indicate uncertainty in an experiment or calculated value. For example, if 15.6 mL is the result of an experiment, then 15 is certain while 6 is uncertain, and the total number of significant figures are 3.

Hence, significant figures are defined as the total number of digits in a number including the last digit that represents the uncertainty of the result.

17.A sample of drinking water was found to be severely contaminated with chloroform, CHCl3, supposed to be carcinogenic in nature. The level of contamination was 15 ppm (by mass).

(i) Express this in percent by mass.
(ii) Determine the molality of chloroform in the water sample.

(i) 1 ppm is equivalent to 1 part out of 1 million (106 ) parts

∴ Mass percent of 15 ppm chloroform in water

=(15/10 ) 100

=1.5 × 10 %

(ii) 100 g of the sample contains 1.5 × 10 -3 g of CHCI3.

⇒1000g of the sample contains 1.5 × 10 -2 g of CHCI3.

∴Molality of chloroform in water 

= 1.5 × 10 -2 g / Molar mass of CHCI3

Molar mass of CHCI3  = 12.00 + 1.00 + 3(35.5)

= 119.5 g mol-1

∴Molality of chloroform in water = 0.0125 × 10 -2 

= 1.25 × 10 -4 m

18.Express the following in the scientific notation:

(i) 0.0048
(ii) 234,000
(iii)8008
(iv)500.0
(v) 6.0012

(i) 0.0048 = 4.8 × 10-3
(ii) 234,000 = 2.34 ×105
(iii)8008 = 8.008 × 103
(iv)500.0 = 5.000 × 102
(v) 6.0012 = 6.0012 × 100

19 .How many significant figures are present in the following

(i) 0.0025
(ii) 208

(iii)5005
(iv)126,000
(v) 500.0
(vi)2.0034

(i) 0.0025
There are 2 significant figures.
(ii) 208
There are 3 significant figures.
(iii)5005
There are 4 significant figures.
(iv)126,000
There are 3 significant figures.
(v) 500.0
There are 4 significant figures.
(vi)2.0034
There are 5 significant figures.

20.Round up the following upto three significant figures.
(i) 34.216
(ii) 10.4107
(iii)0.04597
(iv)2808

(i) 34.2
(ii) 10.4
(iii)0.0460
(iv)2810

21.The following data are obtained when dinitrogen and dioxygen react together to form different compounds:

    Mass of dinitrogen Mass of dioxygen
 i)14 g16 g
ii)14 g32 g
iii)28 g32 g
iv)28 g80 g

(a) Which law of chemical combination is obeyed by the above experimental data? Give its
 statement.
(b) Fill in the blanks in the following conversions:

(i) 1 km = ……………………. mm ……………………. pm
(ii) 1 mg = ……………………. kg ……………………. ng
(iii)1 mL = ……………………. L ……………………. dm3

(a) If we fix the mass of dinitrogen at 28g, then the masses of dioxygen that will combine with the fixed mass of dinitrogenen are 32 g, 64 g, 32 g, and 80 g.
The masses of dioxygen bear a whole number ratio of 1:2:1:5. Hence, the given experimental data obeys the law of multiple proportions. The law states that if two elements combine to form more than one compound, then the masses of one element that combines with the fixed mass of another element are in the ratio of small whole numbers

(b)(i) 1 km = 1 km × ( 1000m/1km  ) ×(100cm/1m) × (10mm /1cm)

∴ 1 km = 10 mm

1 km = 1 km × (1000m/1km )( 1pm/10 -12 m)

∴ 1 km = 1015  pm

Hence, 1 km = 106  mm =1015  pm

(ii) 1 mg = 1 mg  × (1g/1000mg ) × (1kg/1000g)

⇒1 mg = 106kg

∴ 1mg =10-6kg = 106ng

(iii) 1 mL = 1 mL × (1L /1000mL)

⇒ 1mg = 10 -3 L

1 mL = 1cm3 = 1cm3 × ( 1dm × 1dm × 1dm / 10cm ×10cm  × 10cm)

⇒ 1 mL=10 -3 dm 3

∴ 1mL = 10 -3 L = 10 -3 dm3

22.If the speed of light is 3.0 × 108 ms-1 , calculate the distance covered by light in 2.00 ns. 

According to the question:

Time taken to cover the distance = 2.00 ns

=2.00 × 10 -9 s

Speed of light = 3.0 × 10 ms -1

Distance travelled by light in 2.00 ns

= Speed of light × Time taken

=(3.0 × 10  ms-1 ) ((2.00 × 10-9 s)

= 6.00 × 10 -1 m

=0.600m

23.In a reaction

A + B2 →AB2

Identify the limiting reagent, if any in the following reaction mixtures.
(i) 300 atoms of A + 200 molecules of B
(ii) 2 mol A + 3 mol B
(iii)100 atoms of A + 100 molecules of B
(iv)5 mol A+ 2.5 mol B
(v) 2.5 mol A + 5 mol B

A limiting reagent determines the extent of a reaction. It is the reactant which is the first to get consumed during a reaction, thereby causing the reaction to stop and limiting the amount of products formed.
(i) According to the given reaction, 1 atom of A reacts with 1 molecule of B. thus, 200 molecules of B will react with 200 atoms of A, thereby leaving 100 atoms of A unused. Hence, B is the limiting reagent.

(ii) According to the reaction, 1 mol of A reacts with 1 mol of B. thus, 2 mol of A will react with only 2 mol of B. As a result, 1 mol of A will not be consumed. Hence, A is the limiting reagent.

(iii)According of the given reaction, 1 atom of A combines with 1 molecule of B. the mixture is stoichiometric where no limiting reagent is present.

(iv)1 mol of atom A combines with 1 mol of molecule B. Thus, 2.5 mol of B will combine with only 2.5 mol of A. As a result, 2.5 mol of A will be left as such. Hence, B is the limiting reagent.

(v) According to the reaction, 1 mol of atom A combines with 1 mol of molecule B. Thus, 2.5 mol of A will combine with only 2.5 mol of B and the remaining 2.5 mol of B will be left as such. Hence, A is the limiting reagent.

24.Dinitrogen and dihydrogen react with each other to produce ammonia according to the following chemical equation:

N 2(g) + H 2(g)  → 2NH 3(g)

(i) Calculate the mass of ammonia produced if 2.00 × 103 g dinitrogen reacts with 1.00 ×103 g of dihydrogen.
(ii) Will any of the two reactants remain unreacted?
(iii)If yes, which one and what would be its mass?

(i) Balancing the given chemical equation,

N 2(g) + H 2(g)  → 2NH 3(g)

From the equation, 1 mole (28g) of dinitrogen reacts with 3 mole (6g) of dihydrogen to give 2 mole (34g) of ammonia

⇒ 2.00 × 10 3 g of dinitrogen will react with (6g /28g ) 2.00 × 10 g dihydrogen i.e.,

 2.00 × 10g of dinitrogen will react with 428.6 g of dihydrogen.
Given,
Amount of dihydrogen =1.00 × 10 g

Hence, N2 is the limiting reagent.
∴ 28 g of N2 produces 34 g of NH3.

Hence, mass of ammonia produced by 2000 g of N2  =( 34g /28g ) 2000g

=2428.57g

(ii) N2 is the limiting reagent and H2 is the excess reagent. Hence, H2 will remain unreacted.

(iii) Mass of dihydrogen left unreacted = 1.00 × 10 g – 428.6 g =571.4 g

25.How are 0.50 mol Na2CO3 and 0.50 M Na2CO3different?

Molar mass of  Na2CO3= (2 ×23) + 12.00 + (3 ×6) 

=106 g mol  -1

Now, 1 mole of Na2COmeans 106 g of Na2CO3.

∴ 0.5 mol of Na2CO =( 106 g / 1 mole ) × 0.5 mol Na2CO3

= 53 g Na2CO3

⇒0.50 M of Na2CO3= 0.50 mol/L Na2CO3

Hence, 0.50 mol ofNa2COis present in 1 L of water or 53 g Na2CO3 is present in 1 L of water.

26.If ten volumes of dihydrogen gas react with five volumes of dioxygen gas, how many volumes of water vapour would be produced?

Reaction of dihydrogen with dioxygen can be written as:

2 H2(g )+ O2(g) → 2H 2 O(g)

Now, two volumes of dihydrogen react with one volume of dihydrogen to produce two volumes of water Vapour.
Hence, ten volumes of dihydrogen will react with five volumes of dioxygen to produce ten volumes of water Vapour.

27.Convert the following into basic units:

(i) 28.7 pm
(ii) 15.15 pm
(iii)25365 mg

(i) 28.7 pm:
1 pm = 10-12 m
∴28.7 pm = 28.7 × 10-12

=2.87 × 10-11m

(ii) 15.15 pm:

1 pm = 10-12 m

∴15.15 pm = 15.15×   10 -12 m

= 1.515  ×   10 -12 m

(iii) 25365 mg

1 mg =10 -3 g

25365 mg = 2.5365 × 10 4 × 10 -3 g

Since,

1g = 10 -3 g

2.5365 × 101 g = 2.5365  × 10-1 × 10 -3 kg

∴25365 mg = 2.5365× 10 -2 kg

28.Which one of the following will have largest number of atoms?
(i) 1 g Au (s)
(ii) 1 g Na (s)
(iii)1 g Li (s)
(iv)1 g of CI2(g)

(i) 1 g of Au (s) =(1/197  ) mol of Au (s)

= (6.022 × 10 23  /197 ) atoms of Au (s)

=3.06 × 10 21  atoms of Au (s)

(ii) 1 g of Na (s) =1/23 mol of Na (s)

=6.022 × 10 23  /23 atoms of Na (s) 

=26.2  × 10 21  atoms of Na (s) 

(iii) 1 g of Li (s) = 1/7 mol of Li(s) 

=(6.022 ×10 23 /7 )atoms of Li (s) 

= 0.86 ×10 23  atoms of Li (s)

=86.0 ×10 21  atoms of Li (s)

(iv) 1 g of CI2 (g) =1/71  mol of CI2 (g) 

(Molar mass of CI2 molecules = 35.5 × 2 = 71 g  mol -1)

=(6.022 ×10 23 /71 ) atoms of CI2 (g)

=0.0848 × 10 23 atoms of CI2 (g)

=8.48 × 10 23 atoms of CI2 (g)

Hence, 1 g of Li (s) will have the largest number of atoms.

29.Calculate the molarity of a solution of ethanol in water in which the mole fraction of ethanol is 0.040 (assume the density of water to be one).

Mole fraction of C2H5OH = Numberof moles C2H5OH / Numberof moles of solution

0.040 = nC2H5OH   /  nC2H5OH   + n H2O                …….(1)

Number of moles present in 1 L water:

H2O  = 1000g/18 g mol -1

H2O = 55.55mol

Substituting the value of n H2O  in equation (1), 

(nC2H5OH   / nC2H5OH  + 55.55   ) = 0.040

nC2H5OH   = 0.040 nC2H5OH  + (0.040) (55.55)

0.96 n C2H5OH  = 2.222 mole

nC2H5OH =(2.222/ 0.96) mole

nC2H5OH  = 2.314 mole

∴Molarity of solution =2.314 mol / 1L 

=2.314M

30.What will be the mass of one 12C atom in g?

1 mole of carbon atoms = 6.023 ×1023  atoms of carbon

=12 g of carbon

∴Mass of one12C atom = 12 g/ 6.023 ×1023 

=1.993 × 10-23g

31.How many significant figures should be present in the answer of the following calculations?

i) = 0.2856 × 298.15 × 0.112 / 0.5785

ii)5 × 5.364

iii) 0.0125 + 0.7864 +0.0215

i) =0.2856 × 298.15 ×0.112 /0.5785

Least precise number of calculation =0.112

∴Number of significant figures in the answer 

= Number of significant figures in the least precise number =3 

ii) 5 × 5.364

Least precise number of calculation =5.364

∴Number of significant figures in the answer = Number of significant figures in  5.364 =4 

iii) 0.0125 + 0.7864 + 0.0215 

Since the least number of decimal places in each term in four, the number of significant figures in the answer is also 4.

32.Use the data given in the following table to calculate the molar mass of naturally occurring argon isotopes:

IsotopeIsotopic molar massAbundance
36 Ar 35.96755 g mol -10.337%
38Ar37.96272 g mol -1 0.063%
40Ar39.9624 g mol -1 99.600%

Molar mass of argon

=[ (35.96755  × 0.337 /100) +(37.96272  × 0.063 /100 ) +(39.9624 × 90.60/100)]g mol -1

=[0.121 +0.024 +39.802]g mol -1

=39.947 g mol -1

33.Calculate the number of atoms in each of the following 

i) 52 moles of Ar
(ii) 52 u of He
(iii) 52 g of He.

(i) 1 mole of Ar = 6.022 × 10 23atoms of Ar

∴52 mole of Ar =52 × 6.022 × 10 23  atoms of Ar

=3.131 ×10 25atoms of Ar

ii) 1 atom of He = 4 u of the 

or ,

4 u of He = 1 atom of He 

1 u of He =1/4  atom of He 

52u of He =52/4 atom of He 

=13 atoms of He

iii) 4 g of He  = 6.022  × 10 23  atoms of He

∴52 g of He = ( 6.022  × 10 23   ×52 ) / 4    atoms of He

=7.8286  × 10 24   atoms of He

34.A welding fuel gas contains carbon and hydrogen only. Burning a small sample of it in oxygen gives 3.38 g carbon dioxide, 0.690 g water and no other products. A volume of 10.0 L
(measured at STP) of this welding gas is found to weigh 11.6 g. Calculate
(i) empirical formula,
(ii) molar mass of the gas, and
(iii) molecular formula.

(i) 1 mole (44 g) of CO2 contains 12 g carbon.

∴ 3.38 g co2 will contain carbon =(12g/44g ) ×3.38 g

= 0.9217 g

18 g of water contains 2 g of hydrogen.

∴0.690 g of water will contain hydrogen  = (2g/18g ) ×0.090

= 0.0767 g

Since carbon and hydrogen are the only constituents of the compound, the total mass of the compound is:

= 0.9217 g + 0.0767 g

=0.9984 g

∴ Percent of C in the compound = (0.9217g/0.9984g) 100

=92.32%

Percent of H in the compound =(0.0767g/0.9984g ) 100

=7.68%

Moles of carbon in the compound = 92.32 /12.00
=7.69

Moles of hydrogen in the compound =7.68/1 

= 7.68

∴Ration of carbon to hydrogen in the compound = 7.69 : 7.68

=1.1

Hence, the empirical formula of the gas is CH.

ii)Given

, Weight of 10.0L of the gas (at S.T.P) =11.6g

∴ Weight of 22.4 L of gas at STP  =(11.6g /10.0L ) 22.4L

=25.984 g

≈26g

Hence, the molar mass of the gas is 26 g

iii)Empirical formula mass of CH = 12 + 1 = 13 g

n=Molar mass of gas/EmpiricalFormula massof gas

=26g/13g

n=2

∴Molecular formula of gas  =(CH)n

=C2H2

35.Calcium carbonate reacts with aqueous HCl to given CaCl2 and CO2 according to the reaction,CaCO 3(s) + 2 HCl (aq) →CaCl 2(aq)  + CO 2(g) H 2O (l)  What mass of CaCO3 is required to react completely with 25 mL of 0.75 M HCl?

0.75 M of HCI = 0.75 mol of HCl are present in 1 L of water

=[ (0.75mol) ×(36.5 gmol -1 )] HCl is present in 1 L of water

=27.375 g of HCI is present in 1 L of water 

Thus, 1000 mL of solution contains 27.375 g of HCl.

∴ Amount of HCl present in 25 mL of solution

= (27.375g/1000 mL )25 mL

=0.6844g

From the given chemical equation,

CaCO 3(s) + 2 HCl (aq) →CaCl 2(aq)  + CO 2(g) H 2O (l)

2 mol of HCl ( 2 × 36.5 =71 g) react with 1 mol of CaCO3 (100 g).

∴ Amount of CaCO3 that will react with 0.6844 g =( 100/71 )0.6844 g

=0.9639 g

36.Chlorine is prepared in the laboratory by treating manganese dioxide (MnO2) with aqueous hydrochloric acid according to the reaction

4HCl (aq) + MnO2(l) + MnCl 2 (aq) + Cl 2 (g)

How many grams of HCl react with 5.0 g of manganese dioxide?

1 mol [55 + 2 × 16 = 87 g] MnO2 reacts completely with 4 mol [4 × 36.5 = 146 g] of HCl

∴ 5.0 g of MnO2 will react with

=(146g /87g ) 5.0 g

=8.4 g of HCl

Hence, 8.4 g of HCl will react completely with 5.0 g of manganese dioxide.

ABOUT US

NERAM classroom is Tamil nadu’s one of the top NATA and JEE coaching center which made great change in students especially from small towns. Started in 2017,and successfully we have alumni who have been placed under top institutes such as NIT ,SPA ,AU and some other good institutes under Anna universities. NERAM classroom offers highly personalised and effective learning programs for aspirants of competitive exams like JEE, NATA etc . NERAM classroom has become one of the most preferred education platforms across India.

 

WHY WE ARE INCOMPATIBLE ?

Our niche is creating personalised learning experiences for every type of learner. NERAM breaks away from a typical classroom teaching structure and uses technology and customisation to enable students to interact with their tutors.In India, students often feel too shy to clarify their doubts in a crowded coaching class. Many students travel long hours to get to class. NERAM breaks these shackles by liberating learning and teaching from the limits of location.Students are given full freedom to interact with teachers,experts and alumni as per their convenient time.

 

WHY OPT US ?

Traditionally, students rely on brick-and-mortar coaching centers, but companies like NERAM classroom are springing up and more students are supplementing offline courses with online tutoring.
NERAM classroom is an incredibly personalized tutoring platform for you, while you are staying at your home. We have grown leaps and bounds to be the best Online Tuition Website in India with immensely talented Master Teachers, from the most reputed institutions. We provide you year-long structured coaching classes for JEE and NATA entrance exam preparation at affordable tuition fees, with an exclusive session for clearing doubts, ensuring that neither you nor the topics remain unattended. Revision notes and formula sheets are shared with you, for grasping the toughest concepts. Assignments, Regular Homeworks, Subjective & Objective Tests promote your regular practice of the topics.

 

PERSONALISED LEARNING AND TUTORING

Traditionally, students rely on brick-and-mortar coaching centers, but companies like NERAM classroom are springing up and more students are supplementing offline courses with online tutoring.
Personalised Learning is the foundation of our unique programs for students. Using knowledge graphs, the program adapts and creates personalised learning journeys for students.Based on the student's progress, personal learning profiles are created which help analyse strengths and areas of improvement. Each concept gets tagged at different levels of difficulty to create a smooth learning curve which keeps students motivated through any new learning challenge. Whenever errors are made, motivating remedial methods are generated to strengthen and improve the student's learning experience.

 

TECHNOLOGY ENABLED LEARNING

At NERAM, we leverage technology to merge best practices like use of well verified study materials, videos, engaging content and quizzes with the best teachers so that every student across India has access to the best learning and guidance for their future .

KEY FEATURES OF NERAM CLASSROOM

  • INTERACTIVE LEARNING SESSIONS
  • NUMBER OF MOCK TESTS
  • FRIENDLY GUIDE
 

BEST TEACHERS & ENGAGING CONTENT

NERAM classroom Learning Programs provide students a holistic learning experience. Students across regions can access the best teachers.Very few teams like BYJU’S, VEDANTA in India have a great sense for the education sector as well as how technology can be used for effective scaling, NERAM is one such team.
Hours and Hours of Study with no fun, is a bad idea for you, foreseeing the long run. To ensure that motivation is stirred in the best proportion for your clear understanding, a good number of quizzes and Objective tests are organized to impart knowledge. FREE Sample Papers and Important questions are extracted, solved and discussed, ensuring that you are 100% prepared before any exam.
Hours and Hours of Study with no fun, is a bad idea for you, foreseeing the long run. To ensure that motivation is stirred in the best proportion for your clear understanding, a good number of quizzes and Objective tests are organized to impart knowledge. FREE Sample Papers and Important questions are extracted, solved and discussed, ensuring that you are 100% prepared before any exam.

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