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Standard XII

Chemistry

Van't Hoff Factor

A solution of...

Question

A solution of sucrose (molar mass $= 342$ ) is prepared by dissolving 68.4 g in 1000 g of water. Given that the vapour pressure of water at 293 K is 0.023 atm, $K_{f}$ for water = $1.86 K m^{- 1}$ and $K_{b}$ for water = $0.52 K m^{- 1}$ . What is the boiling point of solution?

272

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373.104

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400

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500

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Solution

The correct option is B $373.104$
The boiling point of solution can be calculated by using the relation,
$Δ T_{b} = \frac{K_{b} \times 1000 \times W_{B}}{W_{A} \times M w_{B}}$
Here $K_{b} = 0.52 K m^{- 1}$
$∴ Δ T_{b} = \frac{0.52 \times 1000 \times 68.4}{1000 \times 342} = {0.104}^{o} C$
$∴$ Boiling point of solution = Boiling point of pure water + $Δ T_{b}$
= $373 + 0.104 = 373.104$ K

Hence, option $B$ is correct.

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Similar questions

Q. The boiling point of a solution containing

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A solution of sucrose (molar mass =342) is prepared by dissolving 68.4 g in 1000 g of water. Given that the vapour pressure of water at 293 K is 0.023 atm, Kf for water =1.86 K m−1 and Kb for water =0.52 K m−1. What is the boiling point of solution?

The correct option is B 373.104The boiling point of solution can be calculated by using the relation,ΔTb=Kb×1000×WBWA×MwBHere Kb=0.52K m−1∴ΔTb=0.52×1000×68.41000×342=0.104oC∴ Boiling point of solution = Boiling point of pure water +ΔTb =373+0.104=373.104 KHence, option B is correct.

A solution of sucrose (molar mass $= 342$ ) is prepared by dissolving 68.4 g in 1000 g of water. Given that the vapour pressure of water at 293 K is 0.023 atm, $K_{f}$ for water = $1.86 K m^{- 1}$ and $K_{b}$ for water = $0.52 K m^{- 1}$ . What is the boiling point of solution?