a container containing 320 gram oxygen 280 gram nitrogen 200 gram hydrogen at 273 Kelvin of volume 22.4 lit . Calculate :
• Mole fraction of each component
• Total Pressure
• Partial Pressure of each
Answers
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The partial pressure of nitrogen in mm is 600
Explanation]
Using the equation given by Raoult's law, we get:
p_A=\chi_A\times P_Tp
A =χ A ×PT
p_{N_2}p N2
= partial pressure of N_2N 2
= ?\chi_{N_2}χ N2
= mole fraction of N_2N 2
=P_{T}PT
= total pressure of solution = 1200 mm
Mole fraction of nitrogen is written as:
\chi_{N_2}=\frac{\text{Moles of }N_2}{\text{Total moles}}χN2
= Total moles Moles of N 2
\text{Number of moles of nitrogen}=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{5.6g}{28g/mol}=0.2molNumber of moles of nitrogen=
Molar mass
Given mass
= 28g/mol5.6g
=0.2mol
\text{Number of moles of oxygen}=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{6.4g}{32/mol}=0.2molNumber of moles of oxygen=
Molar mass
Given mass
= 32/mol6.4g
=0.2mol
\chi_{N_2}=\frac{0.2}{0.2+0.2}=0.5χ N 2
= 0.2+0.20.2
=0.5
p_{N_2}=0.5\times 1200=600mmp N 2
=0.5×1200=600mm