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}

 

Answers (1)

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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