Molarity of sample $\times$ volume $H_{2}O_{2}$ is given by the relation  Option 1) $m=\frac{x}{11.2}$ Option 2) $m=\frac{x}{5.6}$ Option 3) $m=\frac{x}{1.12}$ Option 4) $\frac{m}{1}=\frac{x}{34}$

As we learnt in

Decomposition -

It decomposes slowly on exposure to light

- wherein

$2H_{2}O\rightarrow 2HO+O_{2}$

H2O2 undergoes decomposition reaction to give

$H_{2}O_{2}\rightarrow H_{2}O +\frac{1}{2} O_{2}$

$\therefore 1\: \: mol\: \: of\: \: H_{2}O_{2}\rightarrow \frac{1}{2}\: \: mol\: \: O_{2}\rightarrow 11.2\: \: L\: \: of\: \: O_{2}\: \: at\: \:STP$

$\therefore expression \:\: would\: \: be\: \:M\times V= \frac{x}{11.2}$

M = Molarity

V = Volume of Sample

x = Volume of gas libertad at STP

$M=\frac{x}{11.2}$

Option 1)

$m=\frac{x}{11.2}$

This option is correct

Option 2)

$m=\frac{x}{5.6}$

This option is incorrect

Option 3)

$m=\frac{x}{1.12}$

This option is incorrect

Option 4)

$\frac{m}{1}=\frac{x}{34}$

This option is incorrect

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