A mixture of `NO_2` and `N_2O_4` has a vapour density of 38.3 at 300K. What is the number of moles of `NO_2` in 60g of the mixture?

Options

A. 0.15
B. 0.26
C. 1
D. 2.78

Answer

C

Explanation

A=0.25 B=0.75
`P_4^C/P_3^C=7/3implies P_B^0=3/7P_A^0`
`P_T=0.25xxP_A^0+0.75xx3/7P_A^0=P_A^0(1/4+9/28)`
`y_AP_T=P_A^0X_Aimplies y_A=(X0.25)/((1/4+9/28))=(((1//4)/(7+9))/28)=7/16 & y_B=9/16`
`P_T^1=7/16xxP_A^0+9/16xx3/7 P_A^0=P_A^0(7/16+27/(16xx7))`
`y_A^1P_T^1=P_A^0 X_A^1 implies y_A^1 =(7/16)/(((49+27)/(16xx7)))=(7xx16xx7)/(16(49+27))=49/76`
`P^(11)=49/76xxP_A^0 +27/76xx3/7 P_A^0=P_A^0((49xx7+81)/(7xx76))`
`y_A^1 P_T^11=P_A^0x_A^11implies y_A^11=P_A^0(P_A^0xx79//76)/((49xx7+81)/(7xx76))=(49xx7xx76)/(76(49xx7+81))`
`y_A^1=0.8089 ge0.8`
There will be 3 repeated distillation to get `X_A=0.8`