$17.0 g$ of $N H_{3}$ completely vapourises at $- {33.42}^{o} C$ and $1$ bar pressure and the enthalpy change in the process is $23.4 k J {mol}^{- 1}$ . The enthalpy change for the vapourisation of $85 g$ of $N H_{3}$ under the same conditions is $k J$

117

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117.00

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117.0

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Solution

Number of moles of $N H_{3} = 5$
So, required $Δ H = 5 \times 23.4$
$= 117 k J$

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17.0 g of NH3 completely vapourises at −33.42oC and 1 bar pressure and the enthalpy change in the process is 23.4 kJ mol−1. The enthalpy change for the vapourisation of 85 g of NH3 under the same conditions is kJ

Number of moles of NH3=5 So, required ΔH=5×23.4 =117 kJ

$17.0 g$ of $N H_{3}$ completely vapourises at $- {33.42}^{o} C$ and $1$ bar pressure and the enthalpy change in the process is $23.4 k J {mol}^{- 1}$ . The enthalpy change for the vapourisation of $85 g$ of $N H_{3}$ under the same conditions is $k J$

Number of moles of $N H_{3} = 5$
So, required $Δ H = 5 \times 23.4$
$= 117 k J$