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Depression in Freezing Point

How many mole...

Question

How many moles of sucrose should be dissolved in 500 gms of water so as to get a solution which has a difference of $104^{0}$ C between boiling point and freezing point?
Given that : $(K_{f} = 1.86 K k g . m o l^{- 1}, K_{b} = 0.52 K k g . m o l^{- 1})$

1.68

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3.36

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8.4

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0.84

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Solution

The correct option is A 0.84
We know that the boiling point and freezing point of water are $100^{0} C$ and $0^{0} C$ respectively, So

$104^{0} C = {Δ T}_{b} - {Δ T}_{f} + 100^{0} C$

$\Rightarrow 4^{0} C = i K_{b} \times m + i K_{f} \times m = i m (K_{b} + K_{f})$

Now sucrose is a non electrolyte, $i = 1$

$∴$ $m = \frac{4^{0} C}{i (K_{b} + K_{f})} = \frac{4^{0} C}{1 ({0.52}^{0} C . k g / m o l + {1.86}^{0} C . k g / m o l)} = 1.68 m$

Since we have $500 g m$ solvent.

Thus, the moles of sucrose in the water is :

$= \frac{1.68 m o l e s s o l u t e}{K g s o l v e n t} \times 0.500 K g$ solvent

$= 0.840$ moles of sucrose

$∴$ Correct answer is D.

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Similar questions

{103.57}^{o} C

(K_{f} = {1.86}^{o} C / m, K_{b} = {0.52}^{o} C / m)

How many mmol of sucrose should be dissolved in 500 gms of water so as to get a difference of 103.57° between boiling an freezing point?

= 342

100

105^{o}

= 0.51

= 1.86

= 342

100 g m

10^{0} C

(K_{b} = 0.51, K_{f} = 1.86)

= 342 g . {m o l}^{- 1}

68.5 g

1000 g

K_{f}

= 1.86 K k g {m o l}^{- 1}

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