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IB Chemistry: Counting particles by mass: the mole — Practice Questions & Answers
446 practice questions available for this topic — here are 10 with full answers and explanations.
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Practice questions with answers
1
Multiple choice · Easy
What is the approximate value of the Avogadro constant?
$6.02\times 10^{23}\,\text{mol}^{-1}$
$6.02\times 10^{-23}\,\text{mol}^{-1}$
$3.01\times 10^{23}\,\text{mol}^{-1}$
$1.66\times 10^{24}\,\text{mol}^{-1}$
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Why The Avogadro constant is $6.02\times 10^{23}$ particles per mole.
2
Multiple choice · Easy
How many moles are present in $18\,\text{g}$ of water ($M = 18\,\text{g mol}^{-1}$)?
0.5 mol
1 mol
2 mol
18 mol
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Why $n = \dfrac{m}{M} = \dfrac{18}{18} = 1\,\text{mol}$.
3
Multiple choice · Medium
How many molecules are present in $0.50\,\text{mol}$ of a substance?
$3.01\times 10^{23}$
$6.02\times 10^{23}$
$1.20\times 10^{24}$
$0.50$
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Why Number of particles $= n \times N_{A} = 0.50 \times 6.02\times 10^{23} = 3.01\times 10^{23}$.
4
Multiple choice · Medium
A compound contains $40.0\%$ C, $6.7\%$ H and $53.3\%$ O by mass. What is its empirical formula?
$\text{CHO}$
$\text{CH}_{2}\text{O}$
$\text{C}_{2}\text{H}_{4}\text{O}_{2}$
$\text{CH}_{4}\text{O}$
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Why Divide by $A_{r}$: C $40.0/12 = 3.33$; H $6.7/1 = 6.7$; O $53.3/16 = 3.33$. Ratio $1:2:1$, giving $\text{CH}_{2}\text{O}$.
5
Multiple choice · Medium
What is the concentration of a solution made by dissolving $0.50\,\text{mol}$ of solute in $250\,\text{cm}^{3}$ of solution?
$0.50\,\text{mol dm}^{-3}$
$1.0\,\text{mol dm}^{-3}$
$2.0\,\text{mol dm}^{-3}$
$0.125\,\text{mol dm}^{-3}$
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Why $c = \dfrac{n}{V} = \dfrac{0.50}{0.250\,\text{dm}^{3}} = 2.0\,\text{mol dm}^{-3}$.
6
Multiple choice · Hard
A compound has empirical formula $\text{CH}_{2}\text{O}$ and a molar mass of $180\,\text{g mol}^{-1}$. What is its molecular formula?
$\text{CH}_{2}\text{O}$
$\text{C}_{3}\text{H}_{6}\text{O}_{3}$
$\text{C}_{6}\text{H}_{12}\text{O}_{6}$
$\text{C}_{5}\text{H}_{10}\text{O}_{5}$
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Why The empirical formula mass is $12 + 2 + 16 = 30$. $180/30 = 6$, so the molecular formula is $(\text{CH}_{2}\text{O})_{6} = \text{C}_{6}\text{H}_{12}\text{O}_{6}$.
7
Fill in the blank · Easy
The molar mass of carbon dioxide, $\text{CO}_{2}$, is
Check answer
Answer:
44 / 44.0 / 44.01
Why $M = 12 + (2\times 16) = 12 + 32 = 44\,\text{g mol}^{-1}$.
8
Fill in the blank · Medium
The number of moles of atoms in $23\,\text{g}$ of sodium ($A_{r} = 23$) is
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Answer:
1 / 1.0 / one
Why $n = \dfrac{m}{M} = \dfrac{23}{23} = 1\,\text{mol}$.
9
Multiple choice · Medium
A compound is found to contain 0.0199 mol Mn and 0.0398 mol Cl. Its empirical formula is
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Why Dividing by the smallest amount gives Mn:Cl = 1:2, so the empirical formula is MnCl2.
10
Multiple choice · Medium
How many moles of chlorine atoms are present in 1.411 g of chlorine? (Ar of Cl = 35.45)
0.0398 mol
0.0796 mol
50.0 mol
0.0199 mol
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Why n = m / Ar = 1.411 / 35.45 = 0.0398 mol.
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