Question

16 g of oxygen and 3 g of hydrogen are mixed and kept in a vessel at 760 mm of Hg pressure at 0oC. What is
the volume occupied by the mixture

Answer 1

Explanation:

According to ideal gas equation

PV = nRT

P = pressure of the gas = 760 mm Hg = 1atm

( 1atm = 760 mmHg)

V = volume of gas == ?

T = temperature of gas = 0°C = 273 K

R = Gas constant = 0.0821 atm / k mol

Number of moles of oxygen = Given mass / moles mass = 16/32 = 0.5 Moles

Number of moles of hydrogen = given mass / moles mass = 3/2 = 1.5 Moles

Moles of gas = 0.5 + 1.5 = 2.0

== nRt/ p == 2 × 0.0821 / 1atm

=> 44.8 L

Thus the total volume occuping by the mixture will be nearly 44.8 L