Question

The percentage (by weight) of sodium hydroxide in a $1.25$ molal $NaOH$ solution is

• A. $4.76\%$
• B. $1.25\%$
• C. $5\%$
• D. $40\%$

Answer 1

Answer: (A)

$4.76\%$

Answer 2

$1.25$ molal $NaOH$ solution means, $1.25$ moles of $NaOH$ are present in $1000 \,g$ of water

$\therefore$ Weight of $NaOH =1.25 \times 40=50 \,g$

Weight of solution $=1000+50=1050 \,g$
$\%$ (by weight) of $NaOH$
$=\frac{\text { wt. of solute }}{\text { wt. of solution }} \times 100$
$=\frac{50}{1050} \times 100=4.76 \%$