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Q.
Mole fraction of solvent in aqueous solution of $NaOH$ having molality of $3$ is
Some Basic Concepts of Chemistry
Solution:
$m=\frac{1000 \cdot x_{B}}{x_{A} \cdot m_{A}}$
$x_{A}+x_{B}=1$
$\therefore x_{A}=\left(1-x_{B}\right)$
$m=\frac{1000 \cdot x_{B}}{\left(1-x_{B}\right) M_{A}}$
Putting $m=3$
$M _{ A }=18$ because aqueous solution is present
$3 =\frac{1000 \cdot x _{ B }}{\left(1- x _{ B }\right) 18}$
$\Rightarrow 54\left(1- x _{ B }\right)=1000 x _{ B }$
$=54-54 x _{ B }=1000 x _{ B }$
$x _{ B } =\frac{54}{1054}$
$\Rightarrow x _{ B }=0.05$
$\therefore x _{ A }=\left(1- x _{ B }\right)=(1-0.05)=0.95$
