JEE MAIN - Physics (2004 - No. 22)
The electrochemical equivalent of a metal is $${3.35109^{ - 7}}$$ $$kg$$ per Coulomb. The mass of the metal liberated at the cathode when a $$3A$$ current is passed for $$2$$ seconds will be
$$6.6 \times {10^{57}}/kg$$
$$9.9 \times {10^{ - 7}}\,kg$$
$$19.8 \times {10^{ - 7}}\,kg$$
$$1.1 \times {10^{ - 7}}\,kg$$
Explicação
The mass liberated $$m,$$ electrochemical equivalent of a metal $$Z,$$ are related as $$m = Zit$$
$$ \Rightarrow m = 3.3 \times {10^{ - 7}} \times 3 \times 2$$
$$ = 19.8 \times {10^{ - 7}}\,kg$$
$$ \Rightarrow m = 3.3 \times {10^{ - 7}} \times 3 \times 2$$
$$ = 19.8 \times {10^{ - 7}}\,kg$$