When 36.0 g of a solute having the empirical formula $C H_{2} O$ is dissolved in 1.20 kg of water, the solution freezes at $- {0.93}^{o} C$ . What is the molecular formula of the compound?

C_{3} H_{5} O_{2}

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C_{3} H_{4} O_{3}

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C_{2} H_{4} O_{2}

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C_{3} H_{4} O_{2}

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Solution

The correct option is C $C_{2} H_{4} O_{2}$
we have
$Δ T_{f} = K_{f} * m$
0.93 = 1.86 * m
molality = $\frac{m a s s o f s o l u t e}{m o l a r m a s s o f s o l u t e * m a s s o f s o l v e n t}$ = $\frac{36}{x * 1.2}$
we get,
0.93 = 1.86 * $\frac{36}{x * 1.2}$
we get,
molecular weight = 60
we have empirical formula = $C H_{2} O$ $
and empirical formula mass = 30
therefore n= $\frac{60}{30}$ = 2
so, molecular formula becomes = $C_{2} H_{4} O_{2}$

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