Question

What is the molar mass of a gas which occupies 48.7 L at 111oC under 698 torr of pressure and has a mass of 21.5 grams?

Answer 1

Answer:

Molar mass of the gas = 15.15 g/mol

Explanation:

PV = nRT

Where,

P = pressure

n = No. of moles

R = Gas constant

T = Temperature

P = 698 torr, 1 torr = 0.00131579 atm

$698\;torr = 698\times0.00131579 = 0.9184\;atm$

Temperature = 111 °C = 100 + 273.15 = 384.15 K

V = 48.7 L

R = 0.082057 L atm/mol K

Now, PV = nRT

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

$n = \frac{0.9184\times48.7}{0.082057\times384.15}$

          =1.4189 mol

Molar mass = Mass/ No. of  moles

                    = 21.5/1.4189

                    =15.15 g/mol