Question

What will be the value of molality for an aqueous solution of 10% w/w $NaOH?(Na=28,0=16,H=1)$

• A. 5
• B. 2.778
• C. 10
• D. 2.5

Answer 1

The correct option is B 2.778

10% w/w means 10 g $NaOH$ dissolves in 100 g solution or 10 g $NaOH$is present in $(100-10=90g)$ of water.

$\therefore$ Weight of $NaOH=10g$

Mol. wt. of $NaOH=23+16+1=40gmo{l}^{-1}$

Moles of $NaOH=\frac{10}{40}=\frac{1}{4}=0.25$

Weight of solvent (water) = 90 g

$\therefore$ Molality (m) $=\frac{MolesofNaOH\times 1000}{Wt.ofwater\left(ing\right)}$

$=\frac{0.25\times 1000}{90}=2.778$