Question

Relative lowering in vapour pressure (RLVP) of acetone solution containing Nal is 0.2. How much acetone is further needed to be added to the same solution to make the RLVP to 0.1? Initial solution contains 10 mole acetone, assume 100% ionization of Nal and Påcetone = 200 Torr at 300 K.​

Answer 1

Answer:

12.5 mole

Explanation:

this is the answer that i got if you have the answer for this please tell if its correct

Answer 2

Answer:

The answer is 12 mol.

Explanation:

$\frac{delta P}{P_{0} } = x_{solute}$

$x_{solute}$= 0.2

$n_{solvent} / n_{total}$= 0.2

​

=0⋅2× 10 mol= 2mol

Then $n_{solvent}$ =10+2 = 12 moles