The boiling point of n-heptane is 36°C . What will be its molar heat of vaporisation?

Question

The boiling point of n-heptane is 36°C . What will be its molar heat of vaporisation? (A)  27.192 kJ mol-1  (B)  98.304 kJ mol-1  (C)  38.72 kJ mol-1  (D) Data are insufficient to calculate it

Tardigrade Admin · Accepted Answer

According to Troutons law, (Δ Hp/Tb)=88 JK -1 mol -1 ∴ Δ Hv=88 × 309 =27192 J mol -1

Q. The boiling point of $n$ -heptane is $3 6^{\circ} C$ . What will be its molar heat of vaporisation?

$27.192 k J m o l^{- 1}$

$98.304 k J m o l^{- 1}$

$38.72 k J m o l^{- 1}$

Data are insufficient to calculate it

According to Troutons law, $\frac{Δ H _{p}}{T _{b}} = 88 J K^{- 1} m o l^{- 1}$
$∴$ $Δ H_{v} = 88 \times 309$
$= 27192 J m o l^{- 1}$

Q. The boiling point of n-heptane is 36∘C . What will be its molar heat of vaporisation?

27.192kJmol−1

98.304kJmol−1

38.72kJmol−1