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Calculate molality of one litre of solution containing $200 gm {CaBr}_{2}$ . Given density of solution equal to $1.0 gm / ml$ . (Atomic mass of $C a = 40, B r = 80$ )

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$1 m$

$1.4 m$

$1.25 m$

$0.8 m$

answer is B.

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Detailed Solution

$m = \frac{mol}{k g}$

$m = [\frac{\frac{200}{200}}{1000 - 200}] \times 1000 = 1.25$

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