Explanation
Mixture contains 20g $$He$$ and 80g $$S{O}_{4}$$
No of moles of $$He$$ = 20/4 = 5 mol
No of moles of $$S{O}_{4}$$ = 80/64 = 1.25 mol
Mole Fraction of $$He$$ = 5/6.25 = 0.8
And Partial Pressure of $$He$$= 0.8 x 1 bar
= 0.8 bar
Molar mass of water $$=18$$ $$g$$
For $$178.2$$ $$g$$ of $$H_2$$ ,$$n_A = 9.9$$ $$X_B= \cfrac {0.1}{0.1 + 9.9} =0.01$$
Molar mass of Glucuse $$=180$$ $$g$$ $$X_A = 0.99$$
For $$18$$ $$g$$ , $$n_B = 0.1$$
For lowering of vapour pressure ,
$$P = P^o \times A= P^oA(1-X_B) = 760(1-0.01)$$
$$P = 760-7.6$$
$$P= 752.4$$ $$torr $$
It is given that, total pressure of water vapor and nitrogen is
$$ {{P}_{{{N}_{2}}}}+{{P}_{{{H}_{2}}O}}=1\,atm $$
$$ \because {{P}_{{{H}_{2}}O}}=0.4\,atm $$
$$ so, $$
$$ {{P}_{{{N}_{2}}}}=0.6\,atm $$
$$ \because {{P}_{1}}{{V}_{1}}={{P}_{2}}{{V}_{2}} $$
$$ 0.6\times {{V}_{1}}={{P}_{2}}\dfrac{{{V}_{1}}}{3} $$
$$ {{P}_{{{N}_{2}}}}=2.1\,atm $$
$$ {{P}_{{{H}_{2}}O}}=0.4\,atm $$
$$ {{P}_{T}}=2.5\,atm $$
Please disable the adBlock and continue. Thank you.