`500 mL` fo `2 M HCl`, `100 mL` of `2 M H_(2) SO_(4)`, and one gram equivalent of a monoacidic alkali are mixed together. `30mL` of this solution requried `20 mL` of `143 g` `Na_(2) CO_(3). xH_(2)O` in one litre solution. Calculate the water of crystallisation of `Na_(2) CO_(3). xH_(2) O`

Answer

A

Explanation

i. Total mEq of acid `= 500 xx 2 + 100 xx 2 xx 2` (`n` factor)
`= 1000 + 400 = 1400`
Total mEq of monoacid alkali = 1000
Acid left after neutralisation with alkali
`= 1400 - 1000 = 400 mEq`
`N` fo acid left `= (400 mEq)/((500 + 100)mL) = (400)/(600) = 0.6 N`
mEq of acid `-= mEq "of"Na_(2)CO_(3)`
`30 mL xx 0.66 N = 20 mL xx N Na_(2)CO_(3)`
Strength of `Na_(2)CO_(3) x H_(2)O = 143 g L^(-1)`
`M_(Na_(2)CO_(3).xH_(2)O) = ("Strenght")/(Mw) = ((143)/(106 + 18x))`
`M_(Na_(2)CO_(3).xH_(2)O) = ((2 xx 143)/(106 + 18 x))` (`n` factor = 2)
Substitution the value of `N` of `Na_(2)CO_(3).xH_(2)O` in equation (i), we get
`30 xx 0.66 x = 20 xx ((2 xx 143)/(106 + 18 x))`
solve for `x`
`x = 10.16 ~~ 10`
So water of crystallisation of `Na_(2)CO_(3).xH_(2) O = 10`
formula : `Na_(2) CO_(3). 10 H_(2) O`