The molal boiling point constant of water is ${0.573}^{\circ} C k g m o l e^{- 1}$ . When $0.1 m o l e$ of glucose
is dissolved in $1000 g$ of water, the solution boils under atmospheric pressure at:

{100.513}^{\circ} C

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{100.0573}^{\circ} C

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{100.256}^{\circ} C

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{101.025}^{\circ} C

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Solution

The correct option is B ${100.0573}^{\circ} C$
$Δ T_{b} = k_{b} m T_{1} - T_{0} = 0.573 \times 0.1$
$T_{1} = 373.0573 K$
For temperature in $^{o} C$
= $373.0573 K - 273 K$
= ${100.0573}^{o} C$

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The molal boiling point constant of water is 0.573∘Ckgmole−1. When 0.1mole of glucoseis dissolved in 1000g of water, the solution boils under atmospheric pressure at:

The correct option is B 100.0573∘CΔTb=kbmT1−T0=0.573×0.1T1=373.0573KFor temperature in oC= 373.0573K−273K= 100.0573oC

The molal boiling point constant of water is ${0.573}^{\circ} C k g m o l e^{- 1}$ . When $0.1 m o l e$ of glucose
is dissolved in $1000 g$ of water, the solution boils under atmospheric pressure at:

The correct option is B ${100.0573}^{\circ} C$
$Δ T_{b} = k_{b} m T_{1} - T_{0} = 0.573 \times 0.1$
$T_{1} = 373.0573 K$
For temperature in $^{o} C$
= $373.0573 K - 273 K$
= ${100.0573}^{o} C$