The percentage (by weight) of sodium hydroxide in a $1.25$ molal $N a O H$ solution is.

4.76 %

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1.25 %

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5 %

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40 %

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Solution

The correct option is D $4.76 %$
$1.25$ molal NaOH solution means, $1.25$ moles of NaOH are present in $1000$ g of water.
$∴$ Weight of NaOH $= 1.25 \times 40 = 50$ g
Weight of solution $= 1000 + 50 = 1050$ g
$%$ (by weight) of NaOH
$= \frac{w t . o f s o l u t i o n}{w t . o f s o l u t i o n} \times 100$
$= \frac{50}{1050} \times 100 = 4.76 %$ .

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