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)

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

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

Weight of solution $=1000+50=1050g$

$\%$ (by weight) of $NaOH$

$=\frac{\text{ wt. of solute }}{\text{ wt. of solution }}\times 100$

$=\frac{50}{1050}\times 100=4.76\%$