Question

By passing a certain amount of charge through NaCl solution, 9.2 litre of $C{l}_2$ were liberated at STP when the same charge is passed through a nitrate solution of metal M, 7.467g of the metal was deposited. If the specific heat of metal is 0.216 cal/g, the formula of metal nitrate will be:

• A. $M{\left({NO}_3\right)}_3$
• B. $M\left({NO}_3\right)$
• C. $M{\left({NO}_3\right)}_2$
• D. None of these

Answer 1

The correct option is A $M{\left({NO}_3\right)}_3$

$\because$ Specific heat $\times$ atomic mass $\simeq$ 6.4

$\therefore$ Atomic mass of metal $\simeq$ 6.4/0.216 $=29.63$

After electrolysis : Eq. of metal $=$ Eq. of $C{l}_2$

$\frac{m_{metal}}{E_{metal}}=\frac{71\times 9.2}{22.4\times 35.5}=0.82$

($\because$ 22.4 litre $C{l}_2$ at STP weighs 71 g)

or $\frac{7.467\times n}{29.63}=0.82$

$\therefore n=3.25=3$ ($\because n$ is an integer and is charge on cation)

$\therefore$ Metal nitrate is $M{\left({NO}_3\right)}_3$