Question

The energy required to vapourise one mole of benzene at it's boiling point is $31.2kJ$. 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. $\approx 743sec$
• B. $400sec$
• C. $3.12\times {10}^{4}sec$
• D. $\approx 371sec$

Answer 1

The correct option is B $400sec$

Power =100 $\frac{joule}{sec}$

For one mole = (31.2X1000) $\frac{joule}{sec}$

for 100gm $\to$ = (31.2X1000) X $\frac{100}{78.11}$ joule

Equal to 399.4 seconds.