Question

Define molality. How many grams of $N{a}_{2}C{O}_3$ should be dissolved in $250$ grams of water to prepare $0.1m$ solution?

Answer 1

Molality is the number of moles of solute dissolved in 1 kg of solvent.
$\text{Molality}=\frac{\text{number of moles of solute}}{\text{mass of solvent in kg}}$
Let W g of sodium carbonate are dissolved.
The molar mass of $N{a}_{2}C{O}_3=2\left(23\right)+12+3\left(16\right)=46+12+48=106$ g/mol
The number of moles of $N{a}_{2}C{O}_3=\frac{\text{W g}}{\text{106 g/mol}}=\text{0.009434 W mol}$
Mass of water $=\text{250 g}=\text{0.250 kg}$
$\text{0.1 m}=\frac{\text{0.009434 W mol}}{\text{0.250 kg}}$
$W=\text{2.65 g}$
Hence, 2.65 grams of $N{a}_{2}C{O}_3$ should be dissolved in 250 grams of water to prepare 0.1 m solution