Answer: The partial pressure of oxygen in the mixture if the total pressure is 525 mmHg is 310 mm Hg
Explanation:
mass of nitrogen = 37.8 g
mass of oxygen = (100-37.8) g = 62.2 g
Using the equation given by Raoult's law, we get:
![p_A=\chi_A\times P_T](https://tex.z-dn.net/?f=p_A%3D%5Cchi_A%5Ctimes%20P_T)
= partial pressure of
= ?
![\chi_{O_2} = mole fraction of O_2=\frac{\text{Moles of }O_2}{\text{Total moles}}](https://tex.z-dn.net/?f=%5Cchi_%7BO_2%7D%20%3D%20mole%20fraction%20of%20O_2%3D%5Cfrac%7B%5Ctext%7BMoles%20of%20%7DO_2%7D%7B%5Ctext%7BTotal%20moles%7D%7D)
= total pressure of mixture = 525 mmHg
![{\text{Moles of }O_2}=\frac{\text {Given mass}}{\text {Molar mass}}=\frac{62.2g}{32g/mol}=1.94moles](https://tex.z-dn.net/?f=%7B%5Ctext%7BMoles%20of%20%7DO_2%7D%3D%5Cfrac%7B%5Ctext%20%7BGiven%20mass%7D%7D%7B%5Ctext%20%7BMolar%20mass%7D%7D%3D%5Cfrac%7B62.2g%7D%7B32g%2Fmol%7D%3D1.94moles)
![{\text{Moles of }N_2}=\frac{\text {Given mass}}{\text {Molar mass}}=\frac{37.8g}{28g/mol}=1.35moles](https://tex.z-dn.net/?f=%7B%5Ctext%7BMoles%20of%20%7DN_2%7D%3D%5Cfrac%7B%5Ctext%20%7BGiven%20mass%7D%7D%7B%5Ctext%20%7BMolar%20mass%7D%7D%3D%5Cfrac%7B37.8g%7D%7B28g%2Fmol%7D%3D1.35moles)
Total moles = 1.94 + 1.35 = 3.29 moles
![\chi_{O_2}=\frac{1.94}{3.29}=0.59](https://tex.z-dn.net/?f=%5Cchi_%7BO_2%7D%3D%5Cfrac%7B1.94%7D%7B3.29%7D%3D0.59)
![p_{O_2}=\chi_{O_2}\times P_T=0.59\times 525=310mmHg](https://tex.z-dn.net/?f=p_%7BO_2%7D%3D%5Cchi_%7BO_2%7D%5Ctimes%20P_T%3D0.59%5Ctimes%20525%3D310mmHg)
Thus the partial pressure of oxygen in the mixture if the total pressure is 525 mmHg is 310 mm Hg
This might be wrong,but im pretty sure its 0.10 m
Evaporation and transpiration