Fill in the blanks. a. `2.24 L` ammonia at `STP` neutralised `100 mL` of a solution of `H_(2) SO_(4)`. The molarity of acid is…….. b. The equivalent weight of a metal carbonate `0.84 g` of which reacts exactly with `40 mL` of `N//2 H_(2) SO_(4)` is ........ c. `1.575 g`, of hydrated oxalic acid `(COOH)_(2). nH_(2) O` is dissolved in water and the solution is made to `250 mL` On titration, `16.68 mL` of this solution is required for neutralisation of `25 mL` of `N//15 NaOH`. The value of water crystallisation, i.e., `n`, is............ d. `1 mL` of `H_(3) PO_(4)` was diluted to `250 mL`. `25 mL` this solution requried `40.0 mL` of `0.10 N NaOH` for neutralisation using phenolphthanlen as indicator. The specific gravity of acid is.............. The density of 1.48 mass percent calcium hydroxide solution is `1.25 g mL^(-1)`. The volume of `0.1 M HCl` solution required to neutralise `25 mL` of this solution is........

Fill in the blanks. a. `2.24 L` ammonia at `STP` neutralised `100 mL`

1.	Fill in the blanks. a. `2.24 L` ammonia at `STP` neutralised `100 mL` of a solution of `H_(2) SO_(4)`. The molarity of acid is…….. b. The equivalent weight of a metal carbonate `0.84 g` of which reacts exactly with `40 mL` of `N//2 H_(2) SO_(4)` is ........ c. `1.575 g`, of hydrated oxalic acid `(COOH)_(2). nH_(2) O` is dissolved in water and the solution is made to `250 mL` On titration, `16.68 mL` of this solution is required for neutralisation of `25 mL` of `N//15 NaOH`. The value of water crystallisation, i.e., `n`, is............ d. `1 mL` of `H_(3) PO_(4)` was diluted to `250 mL`. `25 mL` this solution requried `40.0 mL` of `0.10 N NaOH` for neutralisation using phenolphthanlen as indicator. The specific gravity of acid is.............. The density of 1.48 mass percent calcium hydroxide solution is `1.25 g mL^(-1)`. The volume of `0.1 M HCl` solution required to neutralise `25 mL` of this solution is........
Answer» Correct Answer - A::B::C::D 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`

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Fill in the blanks. a. The equivalent weight of `NaHCO_(3)` is ……..and of `SO_(2)` is ……… b. 2 mol of 50% pure `Ca(HCO_(3))_(2)` on heating forms 1 mol of `CO_(2)`. The % yield of `CO_(2)` is …….. c. `5 g` of `K_(2) SO_(4)` was dissolved in `250 mL` of solution. The volume of this solution that should be used so that `1.2 g` of `BaSO_(4)` be precipitated fromk `BaCl_(2)` is ....... (molecular mass of `K_(2) SO_(4) = 174` and `BaSO_(4) = 233`) d. The residue obtained on strongly heating `2.76 g Ag_(2) CO_(3)` is ........ `[Ag_(2)CO_(3) overset(Delta)rarr Ag + CO_(2) + O_(2)]` Atomic mass of `Ag = 108`
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