Home
Class 11
CHEMISTRY
A mixture of ZnCl(2) and PbCl(2) can be ...

A mixture of `ZnCl_(2)` and `PbCl_(2)` can be sepqrated by

A

Distillation

B

Crystallization

C

Sublimation

D

Adding acetic acid

Text Solution

Verified by Experts

The correct Answer is:
B

`PbCl_(2)` is insoluble in water, but `ZnCl_(2)` is soluble. So `ZnCl_(2)` can be separated by crystallizing it from the solution.
Promotional Banner

Similar Questions

Explore conceptually related problems

(A) : PbCl_(2) and HgCl_(2) precipitates can be separated by hot water. (R ) : Aqueous solution of PbCl_(2) give yellow precipitate with K_(2)CrO_(4) solution.

Statement 1 : PbCl_(2) and Hg_(2)Cl_(2) precipitates can be separated by hot water . Statement 2 : Hg_(2)Cl_(2) is backend by aq. NH_(3)

When 2-methyl propan-2-ol is treated with a mixture of conc. HCl and ZnCl_(2) , turbidity appears immediately due to the formation of

An aqueous solution is prepared by dissolving a mixture containing ZnCl_(2),CdCl_(2) and CuCl_(2) . Now H_(2)S gas is passed through the aqueous oslution of salt to form precipitate.

Statement-1: Zn cannot be obtained by the electrolysis of mixture of aqueous ZnCl_(2),AgNO_(3) and CuSO_(4) . Statement-2: Standard oxidation potential of Zn is higher than that of Ag and Cu.

A solution is 1 molar in each of NaCl, CdCl_(2), ZnCl_(2) and PbCl_(2) . To this Sn metal is added, which of the following is true? Given E^(@) (Pb^(2+)//Pb = -0.126V) E_(Sn^(2+)//Sn)^(@) = -0.136V, E_(Cd^(2+)//Cd)^(@) = -0.40V E_(Zn^(2+)//Zn)^(@) = -0.763V, E_(Na^(+)//Na)^(@)= -2.71V

What is the role of ZnCl_(2) in dry cell ?

Assertion : PbI_(4) doesn't exist and converts into PbI_(2) and I_(2) spontaneously at room temperature but PbCl_(4) needs heatin to convert into PbCl_(2) and Cl_(2) . Reason : Pb^(2+) is more stable than Pb^(4+) due to inert pair effect.

Zn(s)+2HCl to ZnCl_(2)+H_(2)