Question

If 2.32g of the vapor of a volatile liquid is able to fill a 364-mL flask at 370C and 671mm Hg, what is the molecular weight of the liquid?

Answer 1

Given that mass of vapor of volatile liquid m = 2.32 g

temperature T = 370 ^oC = 370 + 273 K= 643 K

pressure P = 671 mmHg = 671/760 atm = 0.883 atm ( 1 atm = 760 mmHg)

volume of flask V = 364.0 mL = 0.364 L

molar mass of liquid (or) molecular weight of liquid M = ?

Ideal gas equation is

PV = nRT where n = mass m/ molar mass M

R = universal gas constant = 0.0821 L.atm/K/mol

PV = (m/M) R T

M = mRT/ PV

= (2.32 g) (0.0821 L.atm/K/mol) (643 K) / (0.883 atm) (0.364 L)

= 381.05 g/mol

M = 381.05 g/mol

Therefore, molecular weight of liquid = 381.05 g/mol