Fill in the blank.
a. `2.24 L` ammonia at `STP` neutralised `100 mL` of a solution of `H_(2) SO_(4)`. The molarity of acid is……..

Answer

A, B, C, D

Explanation

a. Eq of `NH_(3) =`
`("Volume of a gas at" STP)/("Volume of 1 equivalent of gas") = (2.24)/(22.4) = 0.1 "equivalent"`
mEq of `NH_(3) = 0.1 xx 10^(3) = 100`
mEq of `H_(2)SO_(4) = 100 xx N`
mEq of `H_(2) SO_(4) = mEq "of" NH_(3)`
`100 xx N -= 100`
`N_(H_(2)SO_(4)) = 1`
`M_(H_(2)SO_(4)) = (1)/(2) = 0.5 M`
b. mEq of `MCO_(3) -= mEq "of" H_(2)SO_(4)`
`[0.84)/(Ew(M) + Ew (CO_(3)^(2-))] xx 10^(3) = 40 xx (1)/(2)`
`((0.84)/(E + (60)/(2))) xx 10^(3) = 20`
`((0.84)/(E + 30)) = (20)/(1000)`
`E = 12`
`Ew` of metal carbonate `= 12 + 30 = 42`
`Mw` of `(COOH)_(2), nH_(2) O = 90 + 18 n`
`[{:("Eq of" (COOH)_(2).nH_(2)O),("in" 250 mL "solution"):}] = ((1.575)/(90 + 18 n)) xx 2`
(`n` factor = 2) .....(i)
`16.68 mL xx N = 25 xx (1)/(15)`
`N` (acid) `= 0.099 ~~ 0.1 Eq L^(-1)`
`(0.1 xx 250)/(1000)`
`= (0.1)/(4)` Eq per `250 mL`
Equating equation (i) and (ii), we get
`((2 xx 1.575)/(90 + 18 n)) = (0.1)/(4)`
Solve for `n`, ` n = 2`
Formula : `(COOH)_(2) . 2H_(2) O`
d. mEq `H_(3)PO_(4) mEq "of" NaOH`
`25 mL xx N = 40 mL xx 0.1 N`
`N "of" H_(3) PO_(4) = 0.16`
`N = (W_(2) xx 1000)/(Ew_(2) xx V_(sol)) (Mw H_(3) PO_(4) = 98, Ew = (98)/(3))`
(`n` factor = 3)
`d_(H_(3)PO_(4)) = (W_(2) xx 1000)/(98//3 xx 250), Wt H_(3)PO_(4) mL^(-1) = 1.32 g mL^(-1)`
e. `N_(Ca(OH)_(2)) = (% "by Weight"xx 10 xx d)/(Ew_(2))`
`= (1.48 xx 10 xx 1.025)/(37)`
mEq of `Ca (OH)_(2) -= mEq "of" HCl`
`25 mL xx 0.41 N -= 0.1 xx V_(HCl)`
`V_(HCl) = 10.25 mL`