Question

The energy required to vapourise one mole of benzene at it's boiling point is $31.2kJ/mol$. For how long a $100W$ electric heater has to be operated in order to vaporize a $100g$ sample of benzene at it's boiling temperature?

• A. $\approx 743$ sec
• B. $400$ sec
• C. $3.9\times {10}^4$ sec
• D. $\approx 371$ sec

Answer 1

Answer: (B)

$400$ sec

Boiling point is a temperature at which the vapour pressure of liquid equals atmospheric pressure.

Energy required to vapourise one mole of benzene is 31.2 KJ/mole.

Mole of benzene = 100 gm

Molar mass of benzene = 78 g/mole

Moles of given benzene sample = $\frac{100}{78}$

Energy required to vapourise $\frac{100}{78}$ mole of benzene = $31.2\times \frac{100}{78}KJ$

Power is Energy per unit time.

Power of heater = $100W$

Time required = $\frac{31.2\times 100\times 1000}{78\times 100}$

= $400sec$