A 6.50 molal solution of $K O H (a q)$ has a density of $1.89 g c m^{- 3} .$ The molarity of the solution is….. $m o l d m^{- 3} .$ (Round off to the nearest integer). [Atomic masses: K:39.0u; o:16.0u; H:1.0 u]

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answer is 9.

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Detailed Solution

$\begin{array}{l} m= \frac{1000×M}{(1000×d) - (M× {(GMW)}_{solute})} \\ 6.5= \frac{1000×M}{(1000×1.89) - (M×56)} \\ 6.5 (1890-56M) =1000M \\ 1890-56M=153.8M \\ 209.8M=1890 \\ M=9 \end{array}$

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