[Zhadows]

Hi, I am just curious if someone could check an answer for me:

A) What is the vapor pressure of a solution prepared by dissolving 20.50 g of sodium nitrate in 100.0 g of water at 25 C. The vp of water at 25 C is 23.8 mm Hg.

I calculated the moles of H20 and NaNO3, did the mol fractionation of H2O [5.55] and divided it by 5.55+mol of NaNO3 * 5 because there are 5 units [1xNA+1xN+3xO], then I utilized Raoult's law which is: Psoln=Psolv*Psolv and got a number of 19.56, does this seem right?



Also, would anyone be able to provide me with some advice in solving this one?

B) What is the vapor pressure of a mixture of benzene and toluene is at 25 C in which 1/3 of the molecules are benzene, and what is the mole fraction of each component in the vapor phase? At 25 C, the vp for benzene = 96.0 mm Hg and 30.3 mm Hg for toluene.

I was thinking of approaching this the same way as I did to part A but taking the delta of both values and having that as my answer, but the 1/3 molecules just confused me, any suggestions?

Thank you.

[LQ43]

NaNO3 * 5 because there are 5 units [1xNA+1xN+3xO],

NaNO₃ does not dissociate into 5 units in water -

what are the ions that make up NaNO₃?

B) What is the vapor pressure of a mixture of benzene and toluene is at 25 C in which 1/3 of the molecules are benzene, and what is the mole fraction of each component in the vapor phase? At 25 C, the vp for benzene = 96.0 mm Hg and 30.3 mm Hg for toluene.

P_total = X_aP^o_A  +  X_bP^o_b

X_a , X_b is the mole fraction or (molecule fraction...)