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Elevation in Boiling Point

18g of glucos...

Question

18g of glucose $(C_{6} H_{12} O_{6})$ is dissolved in 1 kg of water in a saucepan. At what temperature will the water boil (at 1 atm)? $K_{b}$ for water is $0.52 K k g m o l^{- 1}$ .

373.202K

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365.202K

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Solution

The correct option is A
373.202K

The given values are:
$W_{s o l u t e} = 18 g$
$W_{s o l v e n t} = 1 k g$
$k_{b} = 0.52 k k g m o l^{- 1}$
First we calculate elevation in the boiling point of solution.
$Δ T_{b} = \frac{K_{b} \times W_{s o l u t e}}{M w_{s o l u t e \times W_{s o l v e n t}}}$
$\frac{0.52 \times 18}{180 \times 1} = 0.052 K$
Since water boils at 373.15 K at 1 atm pressure, therefore the boiling point of solution will be
$T_{B} = T_{b} + Δ T_{b} = 373.15 + 0.052 = 373.202 K$
Thus, the boiling point of solution is 373.202 K .

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18g of glucose (C6H12O6) is dissolved in 1 kg of water in a saucepan. At what temperature will the water boil (at 1 atm)? Kb for water is 0.52Kkgmol−1.

18g of glucose $(C_{6} H_{12} O_{6})$ is dissolved in 1 kg of water in a saucepan. At what temperature will the water boil (at 1 atm)? $K_{b}$ for water is $0.52 K k g m o l^{- 1}$ .