32 g of hydrated magnesium sulphate $M g S O_{4} . x H_{2} O$ , when dissolved in 84 g of water, the solution freezes at $- {4.836}^{0} C$ . If $K_{f} =$ 1.86 K kg mol $^{- 1}$ and $M g S O_{4}$ is a strong electrolyte, what is the value of $x$ ?

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Solution

The correct option is D 7
Molar mass of salt $= 120 + 18 x$
Mass of water present in the salt $= \frac{18 x}{120 + 18 x} \times 32 g$
Molality of the solution $= \frac{32}{120 + 18 x} \times ⎡ ⎢ ⎢ ⎢ ⎣ \frac{1000}{84 + \frac{18 \times 32 x}{120 + 18 x}} ⎤ ⎥ ⎥ ⎥ ⎦$
$= \frac{4000}{1260 + 261 x}$
$Δ T_{f} = 4.836 = 2 \times 1.86 \times \frac{4000}{1260 + 261 x}$
$x = 6.9 \approx 7$

Option A is correct.

Suggest Corrections

Similar questions

M g S O_{4} . x H_{2} O

- {4.836}^{0} C

K_{f} =

m o l^{- 1}

M g S O_{4}

x

6 g

- {0.93}^{0} C

g m o l^{- 1}

(K_{f} o f H_{2} O = 1.86 K K g m o l^{- 1})

40

450

- {0.8}^{\circ} C

K_{f}

1.86

K_{f}

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