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प्रश्न
A 0.15 m aqueous solution of KCl freezes at - 0.510 °C. Calculate i and osmotic pressure at 0 °C. Assume the volume of solution equal to that of water.
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उत्तर
Given:
Molality of solution = m = 0.15 m
Freezing point of solution = Tf = - 0.510 °C
Temperature = 0 °C = 273 K
To find:
1. The value of van’t Hoff factor (i)
2. Osmotic pressure of solution
Formulae:
1. Δ Tf = Kfm
2. i = `(triangle "T"_"f")/(triangle "T"_"f")_0`
3. π = MRT = `("n"_2 "RT")/"V"`
4. i = `pi/pi_0`
Calculation:
`triangle "T"_"f" = "T"_"f"^0 - "T"_"f"`
`= 0^circ "C" - (- 0.510 ^circ "C")` = 0.510 °C = 0.510 K
m = 0.15 m = 0.15 mol kg–1
Now, using formula (i),
Δ Tf = Kfm
`(triangle "T"_"f")_0` = 1.86 K kg mol-1 × 0.15 mol kg-1 = 0.279 K
Now, using formula (ii),
i = `(triangle "T"_"f")/(triangle "T"_"f")_0 = (0.510 "K")/(0.279 "K")` = 1.83
Now, using formula (iii),
(π)0 = MRT
`= "n"_2/"V" "RT"`
`= (0.15 "mol" xx 0.08205 "dm"^3 "atm" * "mol"^-1 "K"^-1 xx 273 "K")/(1 "dm"^3)`
= 3.36 atm
Now, using formula (iv),
i = `pi/pi_0` = 1.83
π = 1.83 × 3.36 atm
π = 6.15 atm
∴ The van’t Hoff factor is 1.83.
∴ The osmotic pressure of solution at 0 °C is 6.15 atm.
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