In a 0.2 molal aqueus solution of a weak acid HX the degree of dissociation is 0.25. The freezing point of the solution will be nearest to: (`K_(f)=1.86 K kg "mol"^(-1)`)
a) `-0.26^(@)C`
b) `0.465^(@)C`
c) `-0.48^(@)C`
d) `-0.465^(@)C`
A
`-0.26^(@)C`
B
`0.465^(@)C`
C
`-0.48^(@)C`
D
`-0.465^(@)C`
Text Solution
Verified by Experts
The correct Answer is:
d
Topper's Solved these Questions
DILUTE SOLUTION
NARENDRA AWASTHI|Exercise Level 1 (Q.62 To Q.91)|1 Videos
0.2 m aqueous solution of a weak acid is 20% dissociated. The boiling point of this solution is (K_(b) of water= 0.52Km^-1)
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) ):
0.5 molal aqueous solution of a weak acid is 20% iionized. If K_f for water is 1.86K kg mol^-1 . The lowering in freezing point of the solution is
0.1 M KI and 0.2 M AgNO_(3) are mixed in 3 : 1 volume ratio. The depression of freezing point of the resulting solution will be [ K_(b)(H_(2)O) = 1.86 K kg "mol"^(-1) ]:
An aqueous solution boils at 101^(@)C . What is the freezing point of the same solution? (Gives : K_(f) = 1.86^(@)C// m "and" K_(b) = 0.51^(@)C//m )
The freezing point of a solution containing 0.1g of K_3[Fe(CN)_6] in 100 g of water. (K_f=1.86 K kg mol^-1) is
If degree of dissociation is 0.01 of decimolar solution of weak acid HA then pK_(a) of acid is :
Calculate depression of freezing point for 0.56 molal aq. Solution of KCl. (Given : K_f(H_(2)O) = 1.8 kg mol^(-1) ).
NARENDRA AWASTHI-DILUTE SOLUTION-Level 3 - Match The Column
