Molar mass of the salt from NaBr,NaNO_3,KI,an | vedclass

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Question

(A) $CaF_2$ reacts with concentrated $H_2SO_4$ to form $HF$ gas,which is colourless.$NaBr + H_2SO_4 \rightarrow NaHSO_4 + HBr$ (followed by oxidation

Vedclass · Accepted Answer

$78$

Vedclass · Answer

(A) $CaF_2$ reacts with concentrated $H_2SO_4$ to form $HF$ gas,which is colourless.$NaBr + H_2SO_4 ightarrow NaHSO_4 + HBr$ (followed by oxidation to reddish-brown $Br_2$ vapours).$NaNO_3 + H_2SO_4 ightarrow NaHSO_4 + HNO_3$ (followed by decomposition to reddish-brown $NO_2$ vapours).$KI + H_2SO_4 ightarrow KHSO_4 + HI$ (followed by oxidation to violet $I_2$ vapours).$CaF_2$ is the only salt that does not evolve coloured vapours.Molar mass of $CaF_2 = 40 + 2 	imes 19 = 40 + 38 = 78 \ g \ mol^{-1}$.

Column-$I$	Column-$II$
$P$. $NaBr + \text{Conc. } H_2SO_4 \rightarrow$	$A$. Colourless gas
$Q$. $Na_2S + \text{dil. } HCl \rightarrow$	$B$. Brown colour
$R$. $NaNO_2 + \text{dil. } HCl \rightarrow$	$C$. Rotten egg smell
$S$. $NaNO_3 + \text{Conc. } H_2SO_4 \rightarrow$	$D$. Paramagnetic

Molar mass of the salt from $NaBr$,$NaNO_3$,$KI$,and $CaF_2$ which does not evolve coloured vapours on heating with concentrated $H_2SO_4$ is . . . . . . $g \ mol^{-1}$. (Molar mass in $g \ mol^{-1}$: $Na: 23$,$N: 14$,$K: 39$,$O: 16$,$Br: 80$,$I: 127$,$F: 19$,$Ca: 40$)

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