Question

Mole fraction of solvent in aqueous solution of $NaOH$ having molality of $3$ is

• A. 0.3
• B. 0.05
• C. 0.7
• D. 0.95

Answer 1

Answer: (D)

$m=\frac{1000\cdot x_B}{x_A\cdot m_A}$

$x_A+x_B=1$
$\therefore x_A=\left(1-x_B\right)$
$m=\frac{1000\cdot x_B}{\left(1-x_B\right)M_A}$
Putting $m=3$
$M_A=18$ because aqueous solution is present
$3=\frac{1000\cdot x_B}{\left(1-x_B\right)18}$
$\Rightarrow 54\left(1-x_B\right)=1000x_B$
$=54-54x_B=1000x_B$
$x_B=\frac{54}{1054}$
$\Rightarrow x_B=0.05$
$\therefore x_A=\left(1-x_B\right)=(1-0.05)=0.95$