A $6.90$ $M K O H$ solution in water has $30 %$ by weight of $K O H$ . The density of $K O H$ solution is:

1.288 g / m L

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12.88 g / m L

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0.1288 g / m L

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none of these

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Solution

The correct option is C $1.288 g / m L$
Here, we have $6.90 M$ solution of $K O H$ . It contains $6.9$ mol of $K O H$ in $1000 c m^{3}$ of solution.

Mass of $6.90$ mol of $K O H = 6.90 \times (39 + 16 + 1) g = 386.4 g$

Again, a $30$ solution contains $30 g$ of $K O H$ in $100 g$ of solution. Therefore, $386.4 g$ of $K O H$ is present in $= \frac{100}{30} \times 386.4 = 1288 g$ of solution

So, density of $K O H$ solution $= \frac{M a s s}{V o l u m e} = \frac{1288}{1000} = 1.288 g / c m^{3}$
Hence, this is the required answer.

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