# NEET Chemistry Hydrogen Questions Solved

A certain sample of hydrogen peroxide is 1.5M solution. It is to be labelled as X volumes. Calculate the value of X.

(A) 16.8 L

(B) 11.8 L

(C) 5.3 L

(D) 6.2 L

or Mass of ${\mathrm{H}}_{2}{\mathrm{O}}_{2}$ per litre=1.5×34=51 g.

51 g of ${\mathrm{H}}_{2}{\mathrm{O}}_{2}$ gives ${\mathrm{O}}_{2}$ at $\mathrm{STP}=\frac{22.4×51}{68}=16.8$

Thus, 1L of ${\mathrm{H}}_{2}{\mathrm{O}}_{2}$ which contains 51 g of ${\mathrm{H}}_{2}{\mathrm{O}}_{2}$ produce ${\mathrm{O}}_{2}$ at STP=16.8 L.

Hence, value of X is =16.8 L

or The given sample is 16.8 volume ${\mathrm{H}}_{2}{\mathrm{O}}_{2}$.

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