Question

Molality of $20$% ($w/w$) aqueous solution of $x$ (M.w.- 50) is:

• A. $2.5m$
• B. $5.0m$
• C. $4.5m$
• D. $6.0m$

Answer 1

Answer: (B)

$5.0m$

$\text{Molality}=\frac{\text{No. of moles of solute}}{\text{Weight of the solvent in kg}}$

Given, $w/w$% of an aqueous solution $=20$%, i.e., $20$ g of solute present in $100$ g of solution.

Hence, mass of the solvent present $=100-20=80$ g

Given, the molecular mass of the solute is $50$.

Number of moles of solute (n) $=\frac{20}{50}=0.4$

Now,
Molality(m) $=\frac{\text{No. of moles}}{\text{Mass of solvent}}$ $=\frac{0.4}{0.08}=5m$