Q9 / 150 · MHT CET 2009 · Chemistry

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

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

Solution

$1.25$ molal NaOH solution means, $1.25$ moles of $\mathrm{NaOH}$ are present in $1000 \mathrm{~g}$ of water $\therefore$ Weight of $\mathrm{NaOH}=1.25 \times 40=50 \mathrm{~g}$ Weight of solution $=1000+50=1050 \mathrm{~g}$ $\%$ (by weight) of $\mathrm{NaOH}$ $$ \begin{array}{l} =\frac{\text { wt. of solute }}{\text { wt. of solution }} \times 100 \\ =\frac{50}{1050} \times 100=4.76 \% \end{array} $$

Hence, the correct answer is option: 'A'.

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