A current is passed through two cells connected in series. The first cell contains X (NO3)3(aq) and the second cell contains Y (NO3)2(aq). The relative atomic mases of X and Y are in the ratio 1: 2. What is the ratio of liberated mass of X to that of Y ?

Question

A current is passed through two cells connected in series. The first cell contains X (NO3)3(aq) and the second cell contains Y (NO3)2(aq). The relative atomic mases of X and Y are in the ratio 1: 2. What is the ratio of liberated mass of X to that of Y ? (A) 3: 2 (B) 1: 2 (C) 1: 3 (D) 3: 1

Tardigrade Admin · Accepted Answer

If atomic mass of X = a, then that of Y = 2 a. As eq. wt = atomic mass/valency Eq. wt. of X = a/3, Eq. wt. of Y = 2 a/2 = a. ( textMass of X/ textMass of Y) = ( textEq.wt, of X/ textEq.wt, of Y) = (a/3/a) = (1/3) = 1:3.

Q. A current is passed through two cells connected in series. The first cell contains $X (N O_{3})$ $_{3 (a q)}$ and the second cell contains $Y (N O_{3})$ $_{2 (a q)}$ . The relative atomic mases of X and Y are in the ratio $1 : 2$ . What is the ratio of liberated mass of $X$ to that of $Y$ ?

3 : 2

1 : 2

1 : 3

3 : 1

2 : 1

If atomic mass of $X = a$ , then that of $Y = 2 a .$
As eq. wt = atomic mass/valency
Eq. wt. of $X = a /3$ , Eq. wt. of $Y = 2 a /2 = a .$
$\frac{Mass of X}{Mass of Y} = \frac{Eq.wt, of X}{Eq.wt, of Y} = \frac{a /3}{a} = \frac{1}{3} = 1 : 3.$

Q. A current is passed through two cells connected in series. The first cell contains X(NO3​)3(aq)​ and the second cell contains Y(NO3​)2(aq)​. The relative atomic mases of X and Y are in the ratio 1:2. What is the ratio of liberated mass of X to that of Y ?