10800C of electricity on passing through the electrolyte solution deposited 2-977 of metal with atomic mass - 106-4gmol - -1 - the charge on the metal cation is-4-2-3-1

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Toppr Admin · Accepted Answer

10800C-x2192-2-977g of metal deposited-96500C-x2192-2-97710800-xD7-96500-26-600gCharge on metal cation is -106-426-6-xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -4

$10800 C$ of electricity on passing through the electrolyte solution deposited $2.977$ of metal with atomic mass ${106.4 g m o l}^{- 1}$ the charge on the metal cation is:
$+ 4$
$+ 2$
$+ 3$
$+ 1$

$10800 C \to 2.977 g$ of metal deposited.
$96500 C \to \frac{2.977}{10800} \times 96500 = 26.600 g$
Charge on metal cation is $= \frac{106.4}{26.6}$
$= + 4$

10800C of electricity on passing through the electrolyte solution deposited 2.977 of metal with atomic mass 106.4gmol−1 the charge on the metal cation is:+4+2+3+1

10800C→2.977g of metal deposited.96500C→2.97710800×96500=26.600gCharge on metal cation is =106.426.6 =+4

$10800 C$ of electricity on passing through the electrolyte solution deposited $2.977$ of metal with atomic mass ${106.4 g m o l}^{- 1}$ the charge on the metal cation is:
$+ 4$
$+ 2$
$+ 3$
$+ 1$

$10800 C \to 2.977 g$ of metal deposited.
$96500 C \to \frac{2.977}{10800} \times 96500 = 26.600 g$
Charge on metal cation is $= \frac{106.4}{26.6}$
$= + 4$