0.32 g of metal gave on treatment with an add 112 mL of hydrogen at NTP . Calculate the equivalent weight of the metal.

Question

0.32 g of metal gave on treatment with an add 112 mL of hydrogen at NTP . Calculate the equivalent weight of the metal. (A)  58  (B)  32  (C)  11.2  (D)  24

Tardigrade Admin · Accepted Answer

Eq. weight of metal =( text weight of metal/ text weight of hydrogen displaced)×1.008 weight of hydrogen liberated =(112/22400)× 2.016 =1.008×10-2 ∴ Eq. weight of metal =(0.32/1.008× 10-2) × 1.008 =32 g equiv-1

Q. $0.32 g$ of metal gave on treatment with an add $112 m L$ of hydrogen at $NTP$ . Calculate the equivalent weight of the metal.

$58$

$32$

$11.2$

$24$

Eq. weight of metal
$= \frac{weight of metal}{weight of hydrogen displaced} \times 1.008$
weight of hydrogen liberated $= \frac{112}{22400} \times 2.016$
$= 1.008 \times 1 0^{- 2}$
$∴$ Eq. weight of metal $= \frac{0.32}{1.008 \times 1 0 ^{- 2}} \times 1.008$
$= 32 g e q u i v^{- 1}$

Q. 0.32g of metal gave on treatment with an add 112mL of hydrogen at NTP . Calculate the equivalent weight of the metal.

58

32

11.2

24