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The freezing point depression of a 0.1 M...

The freezing point depression of a 0.1 M aq. Solution of weak acid (HX) is `-0.20^(@)"C".` What is the value of equilibrium constant for the reaction:
`"HX(aq)"iff"H"^(+)"(aq)"+"X"^(-1)"(aq)"`
[Given: `"K"_("f")" for water=1.8 Kg mol"^(-1)"K. & Molality=Molarity"`]

A

`1.46xx10^(-4)`

B

`1.35xx10^(-3)`

C

`1.21xx10^(-2)`

D

`1.35xx10^(-4)`

Text Solution

Verified by Experts

The correct Answer is:
B

`{:(,HX("aq"),1H^(+)(aq),+, X^(-)(aq)),(t=0,0.1M,0,,0),("At equilibrium",(0.1-"x"),"x",,"x"):}`
`Delta"T"_("f")=(0.1+"x")xx"K"_("b")`
` 0.20=(0.1+"x")xx1.8`
`"K"=("x"xx"x")/((0.1-"x"))=(0.011xx0.011)/((0.1-0.011))`
`"K"=1.35xx10^(-3)`
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