Two volatile substances, A and B are dissolved in one another and the resulting solution has a vapor pressure of 137 torr. If the mole fraction of B is 0.230 and the vapor pressure of pure B is 162 torr, what is the vapor pressure of pure A in torr?

Vapor pressure of pure A is 129.5torr

Explanation:

When two or more liquids produce an ideal solution, the vapour pressure of the mixture follows the equation (Raoult's law):

Where P is vapour pressure and X mole fraction of liquids A and B. Pt is the pressure of the solution

As mole fraction of B is 0.230, mole fraction of A is:

Computing in Raoult's law with the given values:

calculate how many grams of ammonia can be produced from 4.21 grams of

n2 + 3h2 = 2nh3

my answer:

25.0g n2 x (1 mol/ 28.0gn2) x (2 mol nh3/ 1 mol n2) x (17.0g nh3/ 1 molnh3) = 30.36g n2

25.0g h2 x (1 mol h2/ 2.00g h2) x (2 mol nh3/ 3 mol h2) x (17.0g nh3/ 1 mol nh3) = 141.67g h2

lr = n2