Home
Class 12
CHEMISTRY
A sodium salt of ternary acid of molybde...

A sodium salt of ternary acid of molybdenum `(` atomic mass `=96)` has the formula `Na_(2)MoO_(n)`. When an acidified solution of `Na_(2)MoO_(n)` is electrolyzed, `O_(2)` gas is liberated corresponding to a volume of `0.112L` at `STP` and `0.32g` of `Mo` is deposited. Find the formula of salt. .

Text Solution

Verified by Experts

`Na_(2)MoO_(n) rarr 2Na^(o+)+MoO_(n)^(2-)`
If `MoO_(n)^(2-)` ion gets dissociated to `Mo_(x+)` which moves towards cathode, then
`1F=1`Eq of `O_(2)-=1 Eq` of `Mo`
Eq of `O_(2)=(Volume of O_(2)(given)i n L)/(Volume of on e Eq of O_(2)L)`
`(n` factor of `O_(2)=4)`.
`=(0.112L)/(22.4//4)=0.02`
`Eq` of `Mo=0.02=(W_(Mo))/(Ew_(Mo))" "(`Let `n` factor of `Mo=x)`
`:. 0.02 =(0.32)/(96//x)`
`:. x=6`
Reduction at cathode`: Mo^(x+)+6e^(-)rarr Mo`
`:.` Oxidation state of `Mo=6`
In `Na_(2)MoO_(n),` we have
`2(+1)+6+n(-2)=0`
`:. n=4`
Molecular formula of `Na_(2)MoO_(n)impliesNa_(2)MoO_(4)`
Promotional Banner

Similar Questions

Explore conceptually related problems

What volume at STP at ammonia gas will be required to be passed into 30 mL of N H_(2) SO_(4) solution to bring down the acid normality to 0.2 N ?

When a quantity of electricity is passed through CuSO_(4) solution, 0.16 g of copper gets deposited. If the same quantity of electricity is passed through acidulated water, then the volume of H_(2) liberated at STP will be [given : at.wt. of Cu = 64].

Calculate the mass of hydrochloric acid in 200 cm^(3) of 0.2 N solution of it. What volume of this acid solution will react exactly with 25 cm^(3) of 0.14 N solution of sodium hydroxide?

112 mL of hydrogen combines with 56 mL of oxygen of form water. When 224 mL of hydrogen is passes over hand cupric oxide, the cupric oxide loses. 0.160 g of weight. All volumes are measured at STP . Show that the result agrees with the law of constant composition (22.4 L hydrogen and oxygen at STP weigh, respectively, 2g and 32 g )

5L of 0.1 M solution of sodium Carbonate contains 53g of N a 2 C O 3 106 g of N a 2 C O 3 10.6 of N a 2 C O 3 5 × 10 2 millimoles of N a 2 C O 3

What is the volume of CO_2 liberated (in litres) at 1 atmosphere and 0^@C when 10 g of 100% pure calcium carbonate is treated with excess dilute sulphuric acid ? (Atomic mass Ca = 40, C = 12, O = 16)

On dissolving 3.24 g of sulphur in 40 g of benzene,the boiling point of the solution was higher than that of benzene by 0.81K .What is the molecular formula of sulphur? ( K_(b) for benzene = 2.53 K kg mol^(-1) , atomic mass of sulphur = 32 g mol^(-1) ).