Question

The molality of the solution containg $20\%$ w/w solution of $NaOH$ is:

• A. $4.5m$
• B. $6.25m$
• C. $0.3m$
• D. $1m$

Answer 1

Answer: (B)

$6.25m$

Let $x$ be the mass of solution.

Mass of $NaOH=\frac{x\times 20}{100}$

Molar mass of $NaOH=40g$

$\therefore$ Number of moles= $\left(\frac{20x}{100}\right)/40=\frac{x}{200}$

Now mass of solvent in the solution= $x-\frac{20x}{100}=\frac{80x}{100}$

$\therefore$ Molality= $\frac{x/200}{\left(80x/100\right)}\times 1000=6.25M$