Answer:
Vapor pressure for the solution is 21.70 Torr
Explanation:
Vapor pressure lowering → Colligative property
P° = Vapor pressure of pure solvent
P' = Vapor pressure of solution
Xm = Mole fraction of solute → Moles of solute / Total moles
Total moles = Moles of solute + moles of solvent
Formula = P° - P' = P° . Xm . i
Let's determine the moles of solute and solvent
20 g / 58.45 g/mol = 0.342 moles NaCl
150 g / 18 g/mol = 8.333 moles H₂O
Total moles → 8.333 + 0.342 = 8.675 moles
Mole fraction of solute = 0.342 / 8.675 = 0.039
i → NaCl ⇒ Na⁺ + Cl⁻ i = 2 (Van't Hoff factor, number of ions dissolved)
Let's replace the data, that was obtained from the problem.
23.56 Torr - P' = 23.56 Torr . 0.039 . 2
P' = - (23.56 Torr . 0.039 . 2 - 23.56 Torr) → 21.70 Torr
The vapor pressure of the solution if the vapor pressure of water at 25°C is 21.70 Torr.
[tex]P^o - P' = P^{o} . X_m . i[/tex]............(i)
where,
Total moles = Moles of solute + moles of solvent
Number of moles:
It is given by dividing mass over molar mass.
20 g / 58.45 g/mol = 0.342 moles NaCl
150 g / 18 g/mol = 8.333 moles H₂O
Total moles → 8.333 + 0.342 = 8.675 moles
Mole fraction of solute = 0.342 / 8.675 = 0.039
i → NaCl ⇒ Na⁺ + Cl⁻ i = 2 (Van't Hoff factor, number of ions dissolved)
On substituting the values in equation (i):
[tex]23.56 \text{ Torr} - P' = 23.56 \text{ Torr} . 0.039 . 2\\\\P' = - (23.56 \text{ Torr} . 0.039 . 2 - 23.56 \text{ Torr} )\\\\P' = 21.70 \text{ Torr}[/tex]
Thus, the vapor pressure of the solution is 21.70 Torr.
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