By passing a certain amount of cahrge through NaCl solution 9.2 litre of Cl_(2) wre liberated at STP. When the same charge is passed through a nitrate solution of a metal M, 7.467 g of the metal was deposited. If the specific heat of the metal is 0.216 cal g^(-1), what is the formula of metal nitrate?

By passing a certain amount of cahrge through NaCl solution 9.2 litre

1.	By passing a certain amount of cahrge through NaCl solution 9.2 litre of Cl_(2) wre liberated at STP. When the same charge is passed through a nitrate solution of a metal M, 7.467 g of the metal was deposited. If the specific heat of the metal is 0.216 cal g^(-1), what is the formula of metal nitrate?
Answer» SOLUTION :By dulong and Petit's LAW, Sp, heat`xx`Atomic mass=6.4 `therefore`Atomic mass of the metal`=(6.4)/("Sp. Heat")=(6.4)/(0.216)=29.63` SUPPOSE the valency of the metal in 'n'. The equivalent wt. of the metal `=(29.63)/(n)` `22.4L` of `Cl_(2)` have mass=71g `therefore9.2L` of `Cl_(2)` will have mass`=(71)/(22.4)xx9.2=29.161g` Applying faraday's second laaw of electrolysis `("Mass of metal DEPOSITED")/("Mass of "Cl_(2)" Liberated")=("Eq. wt. of metal")/("Eq. at. of chlorine")` `(7.467)/(29.161)=(29.64//n)/(35.5)` or `(29.63)/(n)=(35.5xx7.467)/(29.161)` or `n=(29.63xx29.161)/(35.5xx7.467)=3.25`. Taking the nearest WHOLE number, valency of metal=3. Hence, the formula of metal nitrate will be `M(NO_(3))_(3)`.