The Mole and Stoichiometry: Question 8

Worked solution

Part (a): Converting volume into dm³

There are $1000\ \text{cm}^3$ in $1\ \text{dm}^3$, so divide by $1000$:

$250\ \text{cm}^3 = \frac{250}{1000} = 0.250\ \text{dm}^3$

Part (b): Moles of sodium carbonate needed

Use $\text{moles} = \text{concentration}\times\text{volume}$:

$\text{moles of Na}_2\text{CO}_3 = 0.200\times0.250 = 0.0500\ \text{mol}$

Part (c): Relative formula mass of sodium carbonate

$\text{Na}_2\text{CO}_3$ contains $2$ sodium atoms, $1$ carbon atom and $3$ oxygen atoms:

$M_r(\text{Na}_2\text{CO}_3) = (2\times23)+12+(3\times16) = 46+12+48 = 106$

Part (d): Mass to weigh out

Use $\text{mass} = \text{moles}\times\text{molar mass}$:

$\text{mass of Na}_2\text{CO}_3 = 0.0500\times106 = 5.30\ \text{g}$

Part (e): Concentration in g/dm³

The concentration in $\text{g/dm}^3$ is the mass dissolved in exactly $1\ \text{dm}^3$ of solution. This can be found either by multiplying the concentration in $\text{mol/dm}^3$ by $M_r$, or by scaling up the mass from part (d):

$\text{concentration (g/dm}^3\text{)} = 0.200\times106 = 21.2\ \text{g/dm}^3$

Check using the mass from (d): $5.30\ \text{g}$ dissolved in $0.250\ \text{dm}^3$ gives $5.30\div0.250=21.2\ \text{g/dm}^3$, which agrees.

Final answers

• (a) Volume $=\boxed{0.250}\ \text{dm}^3$
• (b) Moles of $\text{Na}_2\text{CO}_3 = \boxed{0.0500}\ \text{mol}$
• (c) $M_r(\text{Na}_2\text{CO}_3) = 106$
• (d) Mass to weigh out $=\boxed{5.30}\ \text{g}$
• (e) Concentration $=\boxed{21.2}\ \text{g/dm}^3$