What is the volume of `O_(2)` liberated at NTP by complete decomposition of 100 mL of 2M solution of `H_(2)O_(2)` ?

Options

A. 2.24 L
B. 22.4 L
C. 44.8 L
D. 11.2 L

Answer

A

Explanation

Molarity = `("Mass of "H_(2)O_(2)//"Molecular mass of "H_(2)O_(2))/("Volume of solution in litre")`
`("Mass of "H_(2)O_(2)//"Molecular mass of "H_(2)O_(2))/("Volume of solution in litre")`
`(2molL^(-1))=("Mass of "H_(2)O_(2))/((34g mol^(-1))xx(1L))`
`therefore` Mass of `H_(2)O_(2)=(2 mol L^(-1))xx(34 g mol^(-1))xx(1 L)`
= 68 g
1000 mL of solution contains `H_(2)O_(2)=68g`
100 mL of solution contains `H_(2)O_(2)=6.8 g`
Now according to decomposition equation, 68.0 g of `H_(2)O_(2)` at NTP evolve oxygen = 22400 mL
`therefore` 6.8 g of `H_(2)O_(2)` at NTP will evolve oxygen
`=(22400)/(68)xx6.8`
`=2240 mLor 2.24L`