0.71 g of chlorine combines with certain weight of a metal giving 1.11 g of its chloride. The equivalent weight of the metal is

Question

0.71 g of chlorine combines with certain weight of a metal giving 1.11 g of its chloride. The equivalent weight of the metal is (A) 40 (B) 20 (C) 80 (D) none

Tardigrade Admin · Accepted Answer

Let metal chloride be MCl n ∴ equivalents of metal =equivalents of chlorine or (1.11-0.71/E) =(0.71/35.5) ∴ E=20

Q. $0.71 g$ of chlorine combines with certain weight of a metal giving $1.11 g$ of its chloride. The equivalent weight of the metal is

40

20

80

none

Let metal chloride be $MC l_{n}$
$∴$ equivalents of metal
=equivalents of chlorine
or $\frac{1.11 - 0.71}{E} = \frac{0.71}{35.5}$
$∴ E = 20$

Q. 0.71g of chlorine combines with certain weight of a metal giving 1.11g of its chloride. The equivalent weight of the metal is

40

20

80

none

Let metal chloride be MCln​ ∴ equivalents of metal =equivalents of chlorine or E1.11−0.71​=35.50.71​ ∴E=20

Q. $0.71 g$ of chlorine combines with certain weight of a metal giving $1.11 g$ of its chloride. The equivalent weight of the metal is

Let metal chloride be $MC l_{n}$
$∴$ equivalents of metal
=equivalents of chlorine
or $\frac{1.11 - 0.71}{E} = \frac{0.71}{35.5}$
$∴ E = 20$