$H_{2} O_{2}$ solution used for hair bleaching is sold as a solution of approximately 5.0 g $H_{2} O_{2}$ per 100mL of the solution. The molecular mass of $H_{2} O_{2}$ is 34. The molarity of this solution is approximately :

0.15 M

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1.5 M

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3.0 M

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3.4 M

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Solution

The correct option is B $1.5 M$
Man of $H_{2} O_{2} = 5 g$
molar man of $H_{2} O_{2} = 34$
number of mole $= \frac{5}{34} = 0.147$
moraity $\frac{0.147}{100 m . p \times 10^{- 3}} = 1.47 M ≃ 1.5 M$

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