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Standard XII

Chemistry

Properties of Enthalpy

36 mL of pure...

Question

$36 m L$ of pure water takes $100 s e c$ to evaporate from a vessel and heater connected to an electric source which delivers $806 w a t t$ . The $Δ H_{v a p o r i z a t i o n}$ of $H_{2} O$ is

40.3 k J m o l^{- 1}

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43.2 k J m o l^{- 1}

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4.03 k J m o l^{- 1}

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None of these

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Solution

The correct option is A $40.3 k J m o l^{- 1}$
$1 w a t t = 1 J s e c^{- 1} Total heat supplied for 36 mL H_{2} O = 806 \times 100 = 80600 J Δ H_{v a p} = \frac{80600}{36} \times 18 = 40300 J o r 40.3 k J m o l^{- 1}$

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36 mL of pure water takes 100 sec to evaporate from a vessel and heater connected to an electric source which delivers 806 watt. The ΔHvaporization of H2O is

The correct option is A 40.3 kJ mol−11 watt=1 J sec−1Total heat supplied for 36 mL H2O=806×100=80600 JΔHvap=8060036×18=40300 J or 40.3 kJ mol−1

$36 m L$ of pure water takes $100 s e c$ to evaporate from a vessel and heater connected to an electric source which delivers $806 w a t t$ . The $Δ H_{v a p o r i z a t i o n}$ of $H_{2} O$ is

The correct option is A $40.3 k J m o l^{- 1}$
$1 w a t t = 1 J s e c^{- 1} Total heat supplied for 36 mL H_{2} O = 806 \times 100 = 80600 J Δ H_{v a p} = \frac{80600}{36} \times 18 = 40300 J o r 40.3 k J m o l^{- 1}$