Question

The vapor pressure of liquid chloroform, CHCl3, is 100. mm Hg at 283 K. A 0.380 g sample of liquid CHCl3 is placed in a closed, evacuated 380. mL container at a temperature of 283 K.
Assuming that the temperature remains constant, will all of the liquid evaporate? _____yes/no
What will the pressure in the container be when equilibrium is reached? _______mm Hg

Answer 1

Answer:

a

No

b

100 mm Hg

Explanation:

From the question we are told that

The vapor pressure of CHCl3, is $P = 100 \ mmHg = \frac{100}{760}= 0.13156 \ atm$

The temperature of CHCl3 is $T = 283 \ K$

The volume of the container is $V_c = 380mL = 380 *10^{-3}\ L$

The temperature of the container is $T_c = 283 \ K$

The mass of CHCl3 is m = 0.380 g

Generally the number of moles of CHCl3 present before evaporation started is mathematically represented as

$n = \frac{m }{M }$

Here M is the molar mass of CHCl3 with the value $M = 119.38 \ g/mol$

=> $n = \frac{ 0.380 }{119.38 }$

=> $n = 0.00318 \ mols$

Generally the number of moles of CHCl3 gas that evaporated is mathematically represented as

$n_g = \frac{PV}{RT}$

Here R is the gas constant with value $R = 0.08206 L \ atm /mol\cdot K$

So

          $n_g = \frac{0.13156* 380 *10^{-3} }{0.08206 * 283}$

          $n_g = 0.00215 \ mols$

Given that the number of moles of  CHCl3 evaporated is less than the number of moles of CHCl3  initially present , then it mean s that not all the liquid evaporated

At equilibrium the temperature of CHCl3 will be equal to the pressure of  air so the pressure at equilibrium is  100 mmHg