1.

A current is passed through two cells connected in series. The first cell contains X(NO_(3))_(3)(aq) and the second cell contains Y(NO_(3))_(2)(aq). The relative atomic masses of X and Y are in the ratio 1:2. what is the ratio of the liberated mass of X to that of Y?

Answer»

Solution :Suppose atomic mass of X=a
Then atomic mass of Y=2A
EQ. wt. of `X=(a)/(3)`, Eq wt. of `Y=(2a)/(2)=2`(`because` VALENCY of X=3, Valency of Y=2)
Applying Faraday's second LAW of electrolysis:
`("Mass of X")/("Mass of Y")=("Eq. wt. of X")/("Eq. wt. of Y")=(a//3)/(a)=(1)/(3)=1:3`


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