Question

7. Please help is due in 30 mins, will rate, Thank you!

A solution is made of two volatile solutes: Chemical A (with a pure vapor pressure of 80.0 mm Hg) and Chemical B (with a pure vapor pressure of 100.0 mm Hg). The solution has a total vapor pressure of (9.88x10^1) mm Hg. If the solution is known to be made from 3.00 mol of A, how many moles of B must there be in the solution? HINT: The total pressure is the sum of each chemical's partial pressure: Pt = PA + PB. Use PA = XA PA° for both the partial pressure of chemicals A and B.

Answer 1

Given, for chemical A, P_A⁰ = 80 mm Hg, number of moles of A, n_A = 3

for chemical B, P_B⁰ = 100 mm Hg, number of moles of B, n_B  = ?

We know, Partial pressure of A, P_A = X_A x P_A⁰ (X_A = represents the mole fraction of component A)

=> P_A = [3/(3+n_B] x 80 mm Hg ----(i)

Partial pressure of B, P_B = X_B x P_B⁰ (X_B = represents the mole fraction of component B)

=> P_B = [n_B/(3+n_B] x 100 mm Hg ----(i)

Now, Total pressure, P_T = P_A + P_B

=> 98.8 mm Hg = [3/(3+n_B] x 80 mm Hg + [n_B/(3+n_B] x 100 mm Hg

=> 98.8 = [240/(3+n_B)] + [100n_B/(3+n_B)]

=> 98.8 x (3+n_B) = 240 + 100n_B

=> 296.4 + 98.8n_B = 240 + 100n_B

=> 1.2n_B = 56.4

=> n_B = 47 moles

Therefore 47 moles of B must be there.