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$17.0 g$ of $N H_{3}$ completely vaporizes at ${33.42}^{0} C$ and 1 bar pressure and the enthalpy change in the process is $23.4 k J m o l^{- 1}$ . The enthalpy change for the vaporizationof $85 g$ of $N H_{3}$ under the same conditions is ____kJ.

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answer is 117.

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Detailed Solution

${NH}_{3} \to {NH}_{3} (g)$

(1) (g)1
1 mole 1 mole
1 bar 1 bar
$Δ H = 23.4 k j / m o l e n = \frac{85}{17} = 5$

For 5 mole $23.4 \times 5 = 117 k j$

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