The strength of 11.2 volume solution of

Question:

The strength of $11.2$ volume solution of $\mathrm{H}_{2} \mathrm{O}_{2}$ is : [Given that molar mass of $\mathrm{H}=1 \mathrm{~g} \mathrm{~mol}^{-1}$ and $\mathrm{O}=16 \mathrm{~g} \mathrm{~mol}^{-1}$ ]

  1.  $13.6 \%$

  2. $3.4 \%$

  3. $34 \%$

  4. $1.7 \%$


Correct Option: , 2

Solution:

$11.2 \mathrm{~V}$ strength of $\mathrm{H}_{2} \mathrm{O}_{2}$ means,

$11.2 \mathrm{~L}$ of $\mathrm{O}_{2}$ is liberated at STP.

$\mathrm{H}_{2} \mathrm{O}_{2} \longrightarrow \mathrm{H}_{2} \mathrm{O}+\frac{1}{2} \mathrm{O}_{2}$

$11.2 \mathrm{~L}$ of $\mathrm{O}_{2}$ at $\mathrm{STP}=0.5 \mathrm{~mol}$

$\therefore \quad$ No. of moles of $\mathrm{H}_{2} \mathrm{O}_{2}=1 \mathrm{~mol}$

i.e., $1 \mathrm{~L}$ of given $\mathrm{H}_{2} \mathrm{O}_{2}$ solution has 1 mole of $\mathrm{H}_{2} \mathrm{O}_{2}$ (i.e., 34 g)

Strength $=\frac{34}{1000} \times 100=3.4 \%$

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