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 NA = 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 N2 → 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:
C2H6, C6H6, C4 H10,
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:
CH4(g) + 2O2(g) →CO2(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
CH4(g) + 2O2(g)
→CO2(g)
+ 2H2O(l)
Molar masses
Methan = 16g mol-1
Oxygen = 32 g mol-1
From the equation
1 mol CH4
reacts with 2 mol O2
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
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