NIOS Class 12 Chemistry Worksheet 1 Solutions: Atoms, Molecules and Chemical arithmetic

 This post provides clear, step by step solutions to NIOS Class 12 Chemistry Worksheet 1 from the chapter Atoms, Molecules and Chemical Arithmetic.

 

 

This worksheet covers concepts like mole, molar mass, empirical and molecular formulae, percentage to solve numericals for NIOS Class 12 Exams. Each answer is explained in a simple , exam ready manner to  help you understand the methods as well as the concept, making your revision fast and effective. 

 

Atoms, Molecules and Chemical arithmetic Worksheet Solution

 

Q1. The mass of a piece of phosphorus is 99.3g. How many moles of phosphorus are present in it?(The atomic mass of phosphorus is 31 amu).

Answer

We use the mole-mass relationship

 

 

Atomic mass of Phosphorus = 34g mol^-1

 

Given mass of phosphorus =99.3g

 

   

 

Answer n = 3.2 mol

Q2. Calculate the molar mass of each of the following in gmol^-1.

(i)       Sodium Hydroxide (NaOH)

(ii)     Copper sulphate (CuSO₄.5H₂O)

(iii)      Sodium carbonate (Na₂CO₃.10H₂O)

(iv)      Aluminium sulphate (Al₂(SO₄)₃)

Answer

 

Atomic masses : Na=23 , O=16 , H=1 , Cu = 63.5 , S=32 , C=12 , Al =27

(i)     Sodium hydroxide(NaOH)

  Molar mass = 23 +16+1= 40 gmol^-1

 

(ii)       Copper(II)sulphate  (CuSO4.5H2O)

Molar mass of CuSO₄  = 63.5 + 32 + (4×16) = 159.5

Molar mass of 5 H₂O = 5 (2×1+16) =90

 

Total molar mass = 159.5 + 90 =249.5 gmol^-1

(iii)     Sodium carbonate (Na2CO3.10H2O)

 

Molar mass of Na₂CO₃ = (2×23)+12+(3×16)=106

 

Molar mass of 10H₂O= 10 ×18 = 180

 

Total molar mass = 106+180 = 286 gmol^-1

(iv)      Aluminium sulphate (Al2(SO4)3)

 Molar mass = (2× 27)+3 (32+4×16)

                            

                     = 54 + 3× 96

 

                     =  342 gmol^-1

 

Q3. How many moles of CaCO₃will weigh 5 grams.

Answer

We use the mole-mass relationship

 

 

 

 

Molar mass of CaCO₃ (M)= 40 + 12+3×16=100

 

Given mass (m) = 5g

 

   
  = 0.05 mol

 

 

Answer – 0.05 moles of CaCO₃

 

Q4. If you need 1.0 × 10²³ molecules of nitrogen for the reaction N₂ + 3H₂ →2NH₃ then:

(a)       How many mass (in gram) of nitrogen is required?(b)      How many quantities of NH₃ will be formed from 1.0 × 10²³ molecules in the above-mentioned process?(c)
What is the volume of NH₃ gas at STP in (b)?

Answer

Given reaction

 

 N₂ + 3H₂ →2NH₃

 

Required molecules of Nitrogen(N₂) = 1.0×10²³

Avogadro constant N_A = 6.022 × 10²³ mol^-1

(a)      Mass of N₂ required

 

        = 0.166mol

 

 

Mass of N₂ = 28

 

Mass = n × M = 0.166×28= 4.65g

 

Answer – 4.65g of N₂

 

(b)     Quantity of Ammonia (NH₃) formed

      

1 mole N₂  → 2 moles NH₂

 

0.166 mil N_{2 →} 2 × 0.166= 0.332 mol NH₃

 

Answer – 0.332 mol of NH₃

 

(c) Volume of NH₃ at STP

Molar volume at STP= 22.7 Lmol^-1

 

Volume = 0.332 × 22.7 = 7.54L

 

Answer – 7.54L of NH₃

Q5. Write down the empirical formula for the following:

C₂H_6, C₆H₆, C₄ H_10, H₂O₂, KCl

Answer

Empirical formula is the simplest whole number ratio of atom of each element present in a compound.

Given formula	Simplest ratio	Empirical formula
Ethane    C2H6	C:H =2:6 =1:3	CH3
Benzene C6H6	C:H =6:6= 1:1	CH
Butane   C4 H10	C:H =4:10=2:5	C2H5
Hydrogen peroxide  H2O2	H:O =2:2 =1:1	HO
Potassium chloride KCl	Already simplest	KCl

 

Q6. The empirical formula of glucose is CH₂O. whose formula mass is 30 amu. If the molecular mass of glucose is 150 amu then, what is the molecular formula of glucose?

Answer

Empirical formula = Formaldehyde empirical unit (CH₂O)

Empirical formula mass = 30 amu

Molecular mass = 150amu

    =

Molecular formula = Empirical formula × n

  = CH₂O × 5 = C₅H₁₀O₅

Answer = C₅H₁₀O₅

Q7. Write down the percentage of Fe and O for Fe₃O₄ compounds.

Answer

Find percentage composition in Iron oxide (Fe₃O₄)

Atomic masses : Fe= 56 ,  O= 16

Molar mass of  Fe₃O₄

(3×56) + (4×16)  = 168 + 64 = 232gmol^-1

Percentage of  Fe

 Mass of  Fe in 1 mol = 3 × 56 = 168g

  

Percentage of  O

Mass of O in 1 mol = 4 × 16 = 64g

 Fe = 72.41%

O =  27.59%

Q8. A  2.4-gram compound of carbon, hydrogen and oxygen yields 3.52 grams of carbon dioxide (CO₂) and 1.44 grams of water (H₂O). If the molecular mass of the compound is found to be 60 amu then:

(a)       What is the mass of carbon, hydrogen and oxygen in 2.4 g of the compound.

(b)      What is the empirical and molecular formula of the compound?

Answer

Given

On combustion, the compound forms 3.52g of carbon dioxide and 1.44g of water form 2.4g sample.

Atomic masses: C= 12 , H=1 , O=16

(a)      Mass of C,H and O in 2.4 g compound  

Molar mass of CO₂

Molar mass of CO₂ = 44g

12g  C is present in 44 g CO₂

 

Mass of hydrogen from H₂O

Molar mass of H₂O =18g

2g H is present in 18g H₂O

Mass of Oxygen in compound

Mass of O= 2.4- (0.96 + 0.16)= 1.28g

Masses present in compound

C= 0.96g

H= 0.16g

O = 1.28g

(b)      Empirical formula and molecular formula

Convert masses to moles

 

 

Now divide by smallest (0.08)

C;H:O= 1:2:1

Empirical formula = CH₂O

Empirical formula mass = 12 +2 +16 =30

Given molecular mass = 60

 

Molecular formula = ( CH₂O ) × 2 = C₂H₄O₂

Q9. In the following reaction:

 CH_4(g) + 2O_2(g) →CO_{2(g) +} 2H₂O_(l)

(a)      How much mass of oxygen will be required for the complete reaction of 24g CH₄?

(b)     How much mass of CH₄ will be required to react 96 g of oxygen?

Answer

CH_4(g) + 2O_2(g)→CO_{2(g) +} 2H₂O_(l)

Molar masses

Methan = 16g mol^-1

Oxygen = 32 g mol^-1

From the equation

1 mol CH₄ reacts with 2 mol O

So, 16 g CH₄   reacts with 2 ×32=64g  O₂

(a)       Mass of O₂   required for 24 g CH₄

        

(b)      Mass of CH₄ required for 96 g O₂

   64 g  O₂ reacts with 16 g CH₄

    

 

Q10. Industrially caustic soda (NaOH) can be prepared by reacting sodium carbonate (Na₂CO₃) with slaked lime. How many grams of sodium hydroxide (NaOH) will be obtained when 2.0 kg of sodium carbonate (Na₂CO₃) is reacted with calcium hydroxide (Ca (OH)₂).

Answer

 Na₂CO₃ + Ca(OH)₂ → 2NaOH + CaCO₃

Molar masses

Sodium carbonate = 106 gmol^-1

Sodium hydroxide = 40g mol^-1

From the equation

1 mol Na₂CO₃ → 2 mol NaOH

So,

106g  Na₂CO₃ → 2 × 40 =80 g NaOH

Given Na₂CO₃ = 2.0 kg= 2000g

    of NaOH

Quick Revision

   ·     1 mole = 6.022 × 10²³ particle (ions,atoms,molecules)

   ·     n=m\M (moles= mass/molar mass)

   ·     molar volume of gas at STP = 22.7 Lmol^-1

   ·     empirical formula = simplest ratio of atoms

   ·    molecular formula = n ×empirical formula

   ·     always balance the equation before mole calculation

Formulas  Used in the Worksheet

(i)           

(ii)               Number of particles =n ×N_A

(iii)

           

(iv)

               

(v)     

 

Extra Practice Questions

Q1. How many moles are present in 44 g of CO₂?

Q2. Find the molar mass of K₂SO₄.

Q3. What volume will 0.5 mol of O₂ occupy at STP?

Q4. Find empirical formula if C=40%. H=6.7% , O=53.3%.

Q5. How many molecules are present in 9g of water?

Answers

    1. 1 mol

    2. 174 g mol^-1

    3. 11.35 L

    4. CH₂O

   5. 3.01 ×10²³

    ^{Related Topics} 

   ^{1. NIOS Class 12 Worksheet 2 Solutions}