Displaystyle - M - 2- -2elongrightarrow M-space 0-275 g of metal M is deposited the cathode due to passage of 1A of current 965 s- Hence the atomic weight of the metal M is -27-555-0110-013-75

Question

Toppr Admin · Accepted Answer

W-Zit0-275-at-xD7-14-xD7-9652-xD7-965000-550-xD7-100-atomic-xA0-weightHence- atomic weight- 55 g

$M^{2 +} + 2 e ⟶ M . 0.275 g$ of metal M is deposited at the cathode due to passage of $1 A$ of current for $965 s$ . Hence the atomic weight of the metal M is :
$27.5$
$55.0$
$110.0$
$13.75$

$W = Z i t 0.275 = \frac{a t . \times 14 \times 965}{2 \times 96500}$
$0.550 \times 100 = a t o m i c w e i g h t$
Hence, atomic weight= $55$ g

M2++2e⟶M. 0.275g of metal M is deposited at the cathode due to passage of 1A of current for 965s. Hence the atomic weight of the metal M is :27.555.0110.013.75

W=Zit0.275=at.×14×9652×965000.550×100=atomic weightHence, atomic weight= 55 g

$M^{2 +} + 2 e ⟶ M . 0.275 g$ of metal M is deposited at the cathode due to passage of $1 A$ of current for $965 s$ . Hence the atomic weight of the metal M is :
$27.5$
$55.0$
$110.0$
$13.75$

$W = Z i t 0.275 = \frac{a t . \times 14 \times 965}{2 \times 96500}$
$0.550 \times 100 = a t o m i c w e i g h t$
Hence, atomic weight= $55$ g