2.0 g of an elements is reacted with aqueous solution containing KOH and `KNO_(3)` to yield `K_(2)XO_(2)` and `NH_(3).NH_(3)` thus liberated is absorbed in 200 " mL of " 0.05 M `H_(2)SO_(4)`. The excess acid required 10 " mL of " 1.5 M `NaOH` for complete neutralisation Which of the following statements is/are correct?

Options

A. The atomic weight of X is 100 g
B. The equivalent weight of X is 50 g
C. The equivalent weight of X is 25 g
D. The atomic weight of X is 200 g

Answer

A, B

Explanation

`4overset(ɵ)(O)H+2H_(2)O+underset(x=0)(X)tounderset(x-4=-2)(XO_(2)^(2-))+cancel(2e^(-))+4H_(2)O]xx4`
`underline(9H_(2)O+cancel(8e^(-))+NO_(3)^(ɵ)toNH_(3)+3H_(2)O+9overset(o+)(O)H)`
`underline(4X+7overset(o+)(O)H+NO_(3)^(ɵ)to4XO_(2)^(2-)+NH_(3)uarr+2H_(2)O))`
`4X+7KOH+KNO_(3)to4K_(2)XO_(2)+NH_(3)uarr+2H_(2)O`
`m" Eq of "H_(2)SO_(4)` (total) `=200xx0.05xx2` (n factor)
`=20`
`m" Eq of "NaOH=10xx1.5xx1` (n factor) `=15`
`m" Eq of "reacted H_(2)SO_(4)=m" Eq of "NH_(3)`
`=m" mol of "NH_(3)("n factor"=1)`
Using mol concept:`1 m" mol of "NH_(3)=4 m" mol of "X`
`5 m" mol of "NH_(3)=20m" mol of " X`
`m" mol of "X=(2.0)/(Aw)xx10^(3)=20`
`Aw of X=100`
`therefore" Eq of "X=(100)/(2)=50`