Question

A solution of a non-volatile solute in water freezes at −0.30°C. The vapour pressure of pure water at 298 K is 23.51 mm Hg and K_f for water is 1.86 degree/mol. The vapour pressure of rain solution at 298 K is ______ mm Hg.

पर्याय

• 10.97

• 12.97

• 17.86

• 23.44

Answer 1

A solution of a non-volatile solute in water freezes at −0.30°C. The vapour pressure of pure water at 298 K is 23.51 mm Hg and K_f for water is 1.86 degree/mol. The vapour pressure of rain solution at 298 K is 23.44 mm Hg.

Explanation:

We have, ΔT = K_f ​× molality

Given, P⁰ = 23.51 mm of Hg; M = 18; ΔT = 0.3, K_f​ = 1.86 K mol⁻¹

Raoult's Law we get,

⇒ `("P"^0 - "P"_"s")/"P"_"s" = (w xx "M")/("mW")`

⇒ `("P"^0 - "P"_"s")/"P"_"s" = (w xx 1000 xx "M")/("m" xx "W" xx 1000)`

⇒ `("P"^0 - "P"_"s")/"P"_"s" = "molality" xx "M"/1000`

⇒ `("P"^0 - "P"_"s")/"P"_"s" = (Δ"T")/"K"_"f" xx "M"/1000`

⇒ `(23.51 - "P"_"s")/"P"_"s" = 0.3/1.86 xx 18/1000`

So, P_s = 23.44 mm Hg