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

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

MHT CETMHT CET 2009

Solution:

$1.25$ 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 $

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