Three vessels (A, B, C) contain `H_(2)O_(2)` solution. In vessel A , 500 mL of 10 vol `H_(2)O_(2)` is present. 100 mL of 3 vol `H_(2)O_(2)` is present in vessel B. Vessel C is filled with 250 mL of 2M `H_(2)O_(2)`. The weight (in g) of `H_(2)O_(2)` persent in these vessels follows the order

Options

A. `C gt A gt B`
B. `C gt B gt A`
C. `B gt A gt C`
D. `A g B gt C`

Answer

D

Explanation

For vessel A,
10 vol `H_(2)O_(2)` means 1 mL of `H_(2)O_(2)` solution gives 10 mL `O_(2)`.
`{:(2H_(2)O_(2),rarr,2H_(2)O,+,O_(2)),("2 mol",,,,"1 mol"),("= 68 g",,,,"22400 mL at NTP"):}`
`:. 22400` mL `O_(2)` is obtained by 68 g `H_(2)O_(2)`.
`because 10` mL `O_(2)` is obtained by `(68 xx 10)/(22400) = 0.03g H_(2)O_(2)`
Vessel B,
30 vo, `H_(2)O_(2)` means 1 mL of `H_(2)O_(2)` solution gives 30 mL `O_(2)`.
Means, 22400 mL of `O_(2)` is obtained by 68 g `H_(2)O_(2)`.
`because 30` mL `O_(2)` is obtained by `(68 xx 30)/(22400) = 0.09 g H_(2)O_(2)`
Vessel C,
Given, it is filled with 250 mL of 2 M `H_(2)O_(2)`.
Molarity `= ("Number of moles of "H_(2)O_(2))/("volume of sol in L")`
`:. 2 = (x//34)/(250//1000)` or `x = 17 g H_(2)O_(2)`
`:.` The weight (in g) of `H_(2)O_(2)` present in vessels follows order :
`C gt B gt A`