Standard enthalpy of vapourisation for $\ce{CCl4}$ is $30.5\text{ kJ mol}^{-1}$. Heat required for vapourisation of $284\text{ g}$ of $\ce{CCl4}$ at constant temperature is _____ kJ. (Given molar mass in $\text{g mol}^{-1}$; $\ce{C}=12$, $\ce{Cl}=35.5$)

JEE Main · 2024 · 29 Jan 2024 · Shift 2 · Chemical Thermodynamics

Standard enthalpy of vapourisation for

CCl X_{4}

30.5 kJ mol^{- 1}

. Heat required for vapourisation of

284 g

CCl X_{4}

at constant temperature is _____ kJ.
(Given molar mass in

g mol^{- 1}

;

C = 12

Cl = 35.5

)

Answer: 56

Molar mass of

CC l_{4} = 12 + 4 (35.5) = 154 g mo l^{- 1}

.

Moles of

CC l_{4}

in 284 g:

n = \frac{284}{154} = 1.844 mol

Heat required for vaporisation:

Q = n \times Δ H_{vap}^{\circ} = 1.844 \times 30.5 = 56.24 kJ

The heat required is 56 kJ.

JEE Main · 2024 · 29 Jan 2024 · Shift 2 · Integer answer