IB Chemistry SL 1.4.3 Molar Mass

Practise using molar mass, in g mol⁻¹, to connect sample mass with amount of substance. Rearrange n = m/M, retain units throughout and combine the result with balanced-equation…

Exam points

rearrange n = m/M to calculate mass, amount or molar mass with correct units
combine a mass-to-mole conversion with coefficients from a balanced chemical equation
link molar mass to concentration or molar volume in a multi-step calculation

[Maximum number: 1]

Chlorine undergoes many reactions.

2.67 g of manganese(IV) oxide was added to $200.0 \mathrm{~cm}^{3}$ of $2.00 \mathrm{moldm} \mathrm{m}^{-3} \mathrm{HCl}$ .

\mathrm{MnO}_{2}(\mathrm{~s})+4 \mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{Cl}_{2}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})+\mathrm{MnCl}_{2}(\mathrm{aq})

Calculate the amount, in mol, of manganese(IV) oxide added.

$<\frac{2.67 \mathrm{~g}}{86.94 \mathrm{~g} \mathrm{~mol}^{-1}}=>0.0307<\mathrm{mol} \gg$

All question bank results loaded

1.4.3—Molar mass (M) question 1