Home
Class 12
CHEMISTRY
Depression in freezing point of 1.10-mol...

Depression in freezing point of 1.10-molal solution of HF is `0.201^(@)C`. Calculate percentage degree of dissoviation of HF (`K_(f)`=1.856 K kg `mol^(-1)`).

Text Solution

Verified by Experts

The correct Answer is:
8.06%18.67 atm
`DeltaT_(f)=ixxK_(f)xxmori=(DeltaT_(f))/(K_(f)xxm)`
`i=((0.201K))/((1.86" K kg mol"^(-1))(0.10"mol kg"^(-1)))=1.0806.`
`"Degree of dissociation of HF "(alpha)=(i-1)/(n-1)=(1.0806-1)/(2-1)`
=0.0806 = 8.06 %.
Promotional Banner

Similar Questions

Explore conceptually related problems

Depression in freezing point of 0.1 molal solution of HF is -0.201^(@)C . Calculate percentage degree of dissociation of HF. (K_(f)=1.86 K kg mol^(-1)) .

0.15 molal solution of NaCI has freezing point -0.52 ^(@)C Calculate van't Hoff factor . (K_(f) = 1.86 K kg mol^(-1) )

The freezing point of a 1.00 m aqueous solution of HF is found to be -1.91^(@)C . The freezing point constant of water, K_(f) , is 1.86 K kg mol^(-1) . The percentage dissociation of HF at this concentration is:-

In a 0.5 molal solution KCl, KCl is 50% dissociated. The freezing point of solution will be ( K_(f) = 1.86 K kg mol^(-1) ):

A 0.1 molal solution of an acid is 4.5% ionized. Calculate freezing point. (molecular weight of the acid is 300 ). K_(f)="1.86 K mol"^(-1)"kg".

The freezing point of 0.1 molal aq. Solution of K_(X)[Fe(CN)_(6)] is -0.744^(@) C. The molal depression constant of water is 1.86 K-kg mol^(-1) . If the salt behaves as strong electrolyte in water, the value of X is :

What is the depression in freezing point of a solution of non-electrolyte if elevation in boiling point is 0.13K ? (k_(b)=0.52 mol^(-1) kg, k_(f)= 1.86 mol^(-1) kg)