Ideal gas equation problem with solution and practice questions

Learning from this problem: In this following problem you can learn about the application of ideal gas equation, mole concept, unit conversion, and determination of molecular weight. You can learn more different concepts from the similar types of practice problems.

Problem1. 1.46 litre of a gas is collected over water at 30 C and 744 mm of Hg . If the gas weighs 1.98 g and vapour pressure of water at 30 C is 32 mm of Hg. What is the molecular weight of gas?

Solution:

Weight of the gas w = 1.98 g

Volume of the gas V = 1.46 litre

Pressure difference P = 744 – 32 = 712 mm of Hg

= (712/760) atm

Temperature of the gas T = (30 +273) K

= 303 K

Ideal gas follows following equation,

PV = n RT ------------------- (1)

Here, P = pressure , V = Volume, n= number of mole, R = Universal gas constant (0.0821) and

T = Temperature

Number of mole can be defined as follows,

n =( Weight in g/ Molecular weight)= w/M

If we substitute everything in equation 1,

PV = (w/M) RT

(712/760) * 1.47 =( 1.98/ M)* 0.0821*303

M = 35.76 g/ mole

More related Chemistry numerical problems for practice:

1. 1.57 litre of a gas is collected over water at 30 C and 750 mm of Hg . If the gas molecular weight 38 g and vapour pressure of water at 30 C is 32 mm of Hg. What is the density of the gas?

2. Density of the gas is 0.8 g/cm^3. It is filled in the 1.6 litre cylindrical flask at 25 C and 746 mm of Hg. What is the molecular weight of the gas?

3. 1.6 litre of gas is collected over methanol at 30 C and 745 mm of Hg. If the gas weighs 2g and vapour pressure of methanol at 30 C is 88 mm of Hg. What is the mole of the gas?