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

a

No

b

100 mm Hg

Step-by-step explanation:

From the question we are told that

The vapor pressure of CHCl3, is
`P = 100 \ mmHg = (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 = (m )/(M )`

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

=>
`n = ( 0.380 )/(119.38 )`

=>
`n = 0.00318 \ mols`

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

`n_g = (PV)/(RT)`

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

So

`n_g  =  (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