Question

Question #655b9

Answer 1

Answer will be (2) 140 mm of Hg .

Explanation:

We have , A + B -------> AB

Mole fraction (X) of A = 0.8 ; X of B will be = 1 - 0.8 = 0.2

Now , we are given , V.P and of pure A and V.P of solution AB as 70 and 84 mm of Hg respectively .

As it forms an ideal solution , therefore ;

V.P of AB = Partial V.P of A + Partial V.P of B

Partial V.P of any component = X of that component * V.P of that component in pure state

Therefore , substituting the values in
V.P of AB = Partial V.P of A + Partial V.P of B ; we get ,

84 = 70 * 0.8 + V.P of pure B * 0.2

Solving this we get 140 mm of Hg as the V.P of pure B .

Answer 2

I get `"140 mm Hg"`.

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The temperature is just for context.

This question is basically asking you to use Raoult's law for ideal binary mixtures:

`\mathbf(P_j = chi_jP_j^"*")`

where:

• `P_j` is the "non-pure" vapor pressure of the solution (i.e. when there is component `i` in it).
• `P_j^"*"` is the vapor pressure of the pure component j (i.e. the solution with no component `i`).
• `chi_j` is the `\mathbf("mol")` fraction of component `j` in solution. As `chi_j -> 1`, `P_j -> P_j^"*"`.

You have:

• `P_A^"*" = "70 mm Hg"`
• `chi_A = 0.8`
• `P_"tot" = "84 mm Hg"`
• `P_B^"*" = ?`

in an ideal binary mixture of `A` and `B`. But this information is really all you need. Recall Dalton's law of partial pressures:

`\mathbf(P_"tot" = P_A + P_B)`

Now, combine this with Raoult's law (best for ideal binary mixtures!) to get:

`P_"tot" = chi_AP_A^"*" + chi_BP_B^"*"`

Here we can use the fact that `chi_A + chi_B = 1` to get:

`color(blue)(P_"tot") = chi_AP_A^"*" + (1 - chi_A)P_B^"*"`

`= chi_AP_A^"*" + P_B^"*" - chi_AP_B^"*"`

Now you can solve for `P_B^"*"`.

`P_"tot" - chi_AP_A^"*" = P_B^"*"(1 - chi_A)`

`color(blue)(P_B^"*") = (P_"tot" - chi_AP_A^"*")/(1 - chi_A)`

`= (84 - 0.8*(70))/(1 - 0.8)`

`= (84 - 56)/(0.2)`

`= 28*5 = color(blue)("140 mm Hg")`