Question

If 27.1 g of ar(g) occupies a volume of 4.21 l, what volume will 1.29 moles of ne(g) occupy at the same temperature and pressure?

Answer 1

Answer : The volume of neon gas will be, 7.99 liters.

Explanation : Given,

Mass of argon (Ar) gas = 27.1 g

Molar mass of argon = 39.95 g/mole

Volume of argon gas = 4.21 L

Moles of neon (Ne) gas = 1.29 mole

First we have to calculate the moles of argon gas.

$\text{Moles of }Ar=\frac{\text{Mass of }Ar}{\text{Molar mass of }Ar}=\frac{27.1g}{39.95g/mole}=0.68moles$

Now we have to calculate the volume of neon gas.

According to the Avogadro's law, the volume of gas is directly proportional to the number of moles of gas at same pressure and temperature. That means,

$V\propto n$

or,

$\frac{V_1}{V_2}=\frac{n_1}{n_2}$

where,

$V_1$ = volume of argon gas

$V_2$ = volume of neon gas

$n_1$ = number of moles of argon gas

$n_2$ = number of moles of neon gas

Now we put all the given values in this formula, we get

$\frac{4.21L}{V_2}=\frac{0.68mole}{1.29mole}$

$V_2=7.99L$

Therefore, the volume of neon gas will be, 7.99 liters.

Answer 2

N of Ar(g) = 27.1g/39.948g/mol = 0.6783819 mol
V of Ar(g) = 4.21 l
n of Ne(g) = 1.29 mol
V of Ne(g) = V
PV = nRT
P/RT = n1/V1 = n2/V2
V2 = n2 × V1/n1 = 1.29 mol × 4.21 l / 0.6783819 mol
V2 = 8.005668 l