Vidaara.orgClass 11 · Chemistry
CodeVID-C11-01-T1-01
Laws of Chemical Combination & Mole Concept — Practice Assignment
Name: ____________________
Roll No.: __________
Date: ____________
General Instructions
- All questions are compulsory.
- Use atomic masses: H=1, C=12, N=14, O=16, Na=23, Ca=40, S=32.
- Take N_A = 6.022 x 10^23 and molar gas volume at STP = 22.4 L.
- Show all working for full credit.
Section A — Multiple Choice Questions
5 × 1 = 5 marks
1.
One mole of any substance contains:
- A.$6.022\times10^{22}$ particles
- B.$6.022\times10^{23}$ particles
- C.$3.011\times10^{23}$ particles
- D.$12\times10^{23}$ particles
2.
The molar mass of $CaCO_3$ is:
- A.$84\,g\,mol^{-1}$
- B.$100\,g\,mol^{-1}$
- C.$106\,g\,mol^{-1}$
- D.$84.5\,g\,mol^{-1}$
3.
The law of conservation of mass was proposed by:
- A.Proust
- B.Dalton
- C.Lavoisier
- D.Avogadro
4.
The number of moles in $22\,g$ of $CO_2$ is:
- A.$0.25$
- B.$0.5$
- C.$1$
- D.$2$
5.
Equal volumes of gases under the same conditions contain equal numbers of molecules — this is:
- A.Dalton's law
- B.Avogadro's law
- C.Proust's law
- D.Gay-Lussac's law
Section B — Short Answer (2 marks)
3 × 2 = 6 marks
6.
Calculate the number of moles in $4.4\,g$ of $CO_2$.
7.
State the law of multiple proportions with one example.
8.
How many molecules are present in $1.7\,g$ of ammonia, $NH_3$?
Section C — Short Answer (3 marks)
2 × 3 = 6 marks
9.
Calculate the mass of $3.011\times10^{23}$ molecules of glucose ($C_6H_{12}O_6$, M = 180 g/mol).
10.
Two oxides of nitrogen contain 63.6% and 46.7% nitrogen. Verify the law of multiple proportions.
Section D — Long Answer (5 marks)
1 × 5 = 5 marks
11.
A sample contains $25\,g$ of $CaCO_3$. Calculate (a) moles of $CaCO_3$, (b) number of formula units, (c) number of oxygen atoms, and (d) mass of carbon present.
Answer Key
Section A — Multiple Choice Questions
- (B) $6.022\times10^{23}$ particles
- (B) $100\,g\,mol^{-1}$
- (C) Lavoisier
- (B) $0.5$
- (B) Avogadro's law
Section B — Short Answer (2 marks)
- n = 4.4/44 = 0.1 mol.
- When two elements form more than one compound, the masses of one element combining with a fixed mass of the other are in small whole-number ratios. Example: CO and CO2, where oxygen masses per 12 g C are 16 g and 32 g (ratio 1:2).
- Molar mass NH3 = 17; n = 1.7/17 = 0.1 mol; N = 0.1 x 6.022 x 10^23 = 6.022 x 10^22 molecules.
Section C — Short Answer (3 marks)
- n = 3.011 x 10^23 / 6.022 x 10^23 = 0.5 mol; mass = 0.5 x 180 = 90 g.
- Oxide 1: per 63.6 g N there is 36.4 g O, so O/N = 0.572. Oxide 2: per 46.7 g N there is 53.3 g O, so O/N = 1.142. Ratio 0.572 : 1.142 = 1 : 2, a small whole number, confirming the law.
Section D — Long Answer (5 marks)
- (a) M = 100 g/mol, n = 25/100 = 0.25 mol. (b) units = 0.25 x 6.022 x 10^23 = 1.506 x 10^23. (c) each unit has 3 O atoms: O atoms = 3 x 1.506 x 10^23 = 4.517 x 10^23. (d) carbon = 0.25 mol x 12 = 3 g.
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