Question

O2 and SO2 gases are filled in ratio of 1:3 by moles in a closed container of 3L at temp of 27 degC . the partial pressure of O2 is 0.60atm , the concentration of SO2 would be :

ans: 0.036

Answer 1

Concept:

According to Dalton's law

Partial pressure = Total pressure×mole fraction

Explanation :

Given that,

$n_{O_{2} }$ = 1

$n_{SO_{2} }$ = 3

$p_{O_{2} }$ = 0.6

Moles fraction of oxygen

$x_{O_{2} } = \frac{1}{1+3}  = 0.25$

$x_{SO_{2} } = (1-0.25) = 0.75$

Total pressure of oxygen $P_{T} = \frac{0.6}{0.25} = 2.4$

Total pressure of both gases would be same.

Using ideal gas equation

PV = nRT

Number of moles of sulphur dioxide

$n_{SO_{2} } = \frac{2.4*3}{0.0821*300} =  0.3$

Concentration = $\frac{moles}{liter} = \frac{0.3}{3} = 0.1 for 3 moles$

For 1 moles of sulphur dioxide = $\frac{0.1}{3} =  0.033$

Conclusion :

Concentration of sulphur dioxide = 0.033