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Standard XII

Chemistry

Depression in Freezing Point

What weight o...

Question

What weight of glycerol should be added to $600 g$ of water in order to lower its freezing point by $10^{o} C$ ?
Given that: $K_{f} = {1.86}^{o} C . m^{- 1}$ .

496 g

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297 g

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310 g

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426 g

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Solution

The correct option is B $297 g$
Given that,
${Δ T}_{f} = 10^{0} C$

Now,

${Δ T}_{f} = K_{f} \times$ molality

$10 = 1.86 \times \frac{x \times 1000}{92 \times 600}$

$x = \frac{10 \times 600 \times 92}{1.86 \times 1000} = 296.8 g$

Wt. of glycerol taken $\approx 297 g$

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What weight of glycerol should be added to 600g of water in order to lower its freezing point by 10oC?Given that: Kf=1.86oC.m−1.

The correct option is B 297gGiven that,ΔTf=100CNow,ΔTf=Kf× molality10=1.86×x×100092×600x=10×600×921.86×1000=296.8gWt. of glycerol taken ≈ 297g

What weight of glycerol should be added to $600 g$ of water in order to lower its freezing point by $10^{o} C$ ?
Given that: $K_{f} = {1.86}^{o} C . m^{- 1}$ .

The correct option is B $297 g$
Given that,
${Δ T}_{f} = 10^{0} C$

Now,

${Δ T}_{f} = K_{f} \times$ molality

$10 = 1.86 \times \frac{x \times 1000}{92 \times 600}$

$x = \frac{10 \times 600 \times 92}{1.86 \times 1000} = 296.8 g$

Wt. of glycerol taken $\approx 297 g$