Question

What is the molar mass of 37.96 g of gas exerting a pressure of 3.29 on the walls of a 4.60 L container at 375 K?

Answer 1

The molar mass of the gas is 77.20 gm/mole.

Explanation:

The data given is:

P = 3.29 atm,   V= 4.60 L   T= 375 K  mass of the gas = 37.96 grams

Using the ideal Gas Law will give the number of moles of the gas. The formula is

PV= nRT    (where R = Universal Gas Constant 0.08206 L.atm/ K mole

Also number of moles is not given so applying the formula

n= mass ÷ molar mass of one mole of the gas.

n = m ÷ x   ( x  molar mass) ( m mass given)

Now putting the values in Ideal Gas Law equation

PV = m ÷ x RT

3.29 × 4.60 = 37.96/x × 0.08206 × 375

15.134 = 1168.1241  ÷ x

15.134x = 1168.1241

x = 1168.1241 ÷ 15.13

x = 77.20 gm/mol

If all the units in the formula are put will get cancel only grams/mole will be there. Molecular weight is given by gm/mole.