Home
Class 12
CHEMISTRY
A 0.5% aqueous solution of KCl was found...

A `0.5%` aqueous solution of `KCl` was found to freeze at `-0.24^(@)C`. Calculate the Van,t Hoff factor and degree of dissociation of the solute at this concentration. (`K_(f)` for water =`1.86 K kg mol`^(-1)`)

Text Solution

Verified by Experts

`DeltaT_(f)=ixxk_(f)xx(w_(2)xx1000)/(Mw_(2)xxW_(1))`
`0.24=ixx1.86xx(0.5xx1000)/(74.5xx99.5) rArr i=1.92`

Total number of particles =`1-alpha+alpha+alpha=1+alpha`
`i=1+a`
`1.92=1+alpha`
So, `alpha=1.92-1=0.92`
i.e.,`92% dissociated`.
Promotional Banner

Similar Questions

Explore conceptually related problems

A 0.5 percent aqueous solution of KCl was found to freeze at 272.76K. Calculate the Van.t Hoff factor and degree of dissociation of the solute at this concentration (K_(f)" for water = 1.86 k.kg.mol"^(-1)) . Normal molar mass of KCl = 74.5.

0.6 mL of acetic acid (CH_(3)COOH) having density 1.06 g mL^(-1) is dissolved in 1 L of water. The depression in freezing point observed for this strength of acid was 0.0205^(@)C .Calculate the Van't Hoff factor and dissociation constant of the acid. K_(f) for H_(2)O=1.86 K kg ^(-1) "mol"^(-1))

A 1.17% solution of NaCl is isotonic with 7.2% solution of glucose. Calculate the Van't Hoff factor of NaCl .

Calculate the amount of KCl which must be added to 1 kg of water so that the freezing point is depressed by 2 K. (K_(f) " for water = " 1.86 " K kg "mol^(-1))

The freezing point of a 0.08 molal solution of NaHSO^(4) is -0.372^(@)C . Calculate the dissociation constant for the reaction. K_(f) for water = 1.86 K m^(-1)

An aqueous solution containing an ionic salt having molality equal to 0.19 freezes at -0.704^(@)C . The Van't Hoff factor of the ionic salt is ( K_(f) for water= 1.86 K m^(-1) )

The degree of dissociation for PtCl_(4) complex is 70% . Calculate the Van't Hoff factor.

15 g of an unknown molecular substance was dissolved in 450 g of water. The resulting solution freezes at - 0.34^(@) C. what is molar mass of the substance ? (K_(f)" for water " = 1.86 k kg " mol"^(-1))