$\begin{array}{1 1}(a)\;34.76g&(b)\;12.38g\\(c)\;1.238g&(d)\;3.476g\end{array}$

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Step 1:

Equivalent weight of ferrous ammonium sulphate =392

$\therefore$ normality of salt solution =$\large\frac{3.92}{392}\times \frac{1000}{100}$

$\Rightarrow 0.1N$

$V_1N_1=V_2N_2$

$V_1N_1\Rightarrow $Mohr's salt

$V_2N_2\Rightarrow KmnO_4$

$20\times 0.1=18\times N_2$

$N_2=\large\frac{20\times 0.1}{18}$

$\Rightarrow \large\frac{1}{9}$$N$

Step 2:

Equivalent weight of $KMno_4=31.6$

$\therefore$ weight of $KMnO_4=31.6$

$\therefore$ Weight of $KMnO_4$ in litre=$31.6\times \large\frac{1}{9}$

$\Rightarrow 3.5$g

Hence (d) is the correct option.

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