Question

The mole fraction of the solute in the 12 molal solution of $N{a}_{2}C{O}_3$ in water as solvent is :

• A. 0.822
• B. 0.177
• C. 1.77
• D. 0.0177

Answer 1

The correct option is B 0.177

Molality of $N{a}_{2}C{O}_3$ $\left(m_{N{a}_{2}C{O}_3}\right)=12m$
This implies that 12 mol of $N{a}_{2}C{O}_3$ are present in $1000g$ of solvent i.e. water.
Moles = $\frac{Mass}{Molarmass}$
Moles of water $\left(n_{H_{2}O}\right)=\frac{1000}{18}=55.55$

$N{a}_{2}C{O}_3$ is solute here.

Mole fraction $\left(\chi \right)$ of $N{a}_{2}C{O}_3$ = $\frac{MolesofN{a}_{2}C{O}_3}{Totalmoles}$
${\chi }_{N{a}_{2}C{O}_3}=\frac{12}{12+55.55}=0.177$