Question

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)`

Answer 1

Correct Answer - 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)`