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Calculate the change in enthalpy \mathrm{(\Delta H)} when \mathrm{50.0 \mathrm{~g}} of water at \mathrm{100^{\circ} \mathrm{C}} is converted to steam at \mathrm{100^{\circ} \mathrm{C}.} The heat of

vaporization of water is \mathrm{40.79 \mathrm{~kJ} / \mathrm{mol}.}

Option: 1

113.1 kJ


Option: 2

-113.1 kJ


Option: 3

11.3 J


Option: 4

-113.J


Answers (1)

best_answer

Step 1: Calculate moles of water (n)

\mathrm{ n=\frac{\text { mass }}{\text { molar mass }}=\frac{50.0 \mathrm{~g}}{18.015 \mathrm{~g} / \mathrm{mol}} \approx 2.774 \mathrm{~mol} }

Step 2: Calculate \mathrm{\Delta H} using the formula

\mathrm{ \Delta H=n \cdot \Delta H_{\text {vap }}=2.774 \mathrm{~mol} \cdot 40.79 \mathrm{~kJ} / \mathrm{mol} \approx 113.1 \mathrm{~kJ} }

Therefore, the change in enthalpy \mathrm{(\Delta H)} is approximately \mathrm{113.1 \mathrm{~kJ}}.

so, the correct option is 1

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chirag

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