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NCERT Exemplar · Q43

Q.The enthalpy of vapourisation of CCl4 is 30.5 kJ mol^-1. Calculate the heat required for the vapourisation of 284 g of CCl4 at constant pressure. (Molar mass of CCl4 = 154 g mol^-1).

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The heat required at constant pressure equals the enthalpy change for the process. Using the given enthalpy of vaporisation and the number of moles in 284 g of CCl₄, the answer is 56.2 kJ.

The key idea here is that enthalpy of vaporisation (ΔHvap\Delta H_{\text{vap}}) is the heat absorbed when one mole of a substance vaporises at constant pressure. Since the problem asks for heat at constant pressure, that heat is the enthalpy change — no extra conversion needed. The only work is to find how many moles are in 284 g and multiply.

  1. Find the number of moles. Molar mass of CCl₄ is given as 154 g mol⁻¹.

n=massmolar mass=284154n = \frac{\text{mass}}{\text{molar mass}} = \frac{284}{154}

Simplify:

n=284154=14277≈1.844 moln = \frac{284}{154} = \frac{142}{77} \approx 1.844 \text{ mol}

(You can keep it as a fraction for exactness — we’ll use the decimal for clarity.)

  1. Apply the enthalpy of vaporisation. ΔHvap=30.5 kJ mol−1\Delta H_{\text{vap}} = 30.5 \text{ kJ mol}^{-1} means each mole requires 30.5 kJ of heat at constant pressure. So total heat qpq_p is:

qp=n×ΔHvap=1.844×30.5q_p = n \times \Delta H_{\text{vap}} = 1.844 \times 30.5

Compute:

1.844×30.5=1.844×(30+0.5)=55.32+0.922=56.242 kJ1.844 \times 30.5 = 1.844 \times (30 + 0.5) = 55.32 + 0.922 = 56.242 \text{ kJ}

Rounding to three significant figures (since 30.5 has three and 284 has three): …

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