Question

Question #bee0c

Answer 1

Volume`rArr`627mL

Explanation:

Volume of `H_2O_2` solution: 24.75mL.

Mass of `H_2O_2` solution`rArr` `rho` `xx`V`=` 24.75gm .

So lets find the moles of `H_2O_2` present in 24.75 gm solution

Given the solution contained 3.85% `H_2O_2` by mass`rArr`

100gm of the solution contain 3.85 gm of `H_2O_2` by mass

24.75gm solution contains`rarr` 3.85`xx(24.75)/(100)`

`color(red)rArr`0.9528 grams of `H_2O_2`

No. of Moles`=(0.9528)/(34)` `rArr`0.0280

Molar volume of a gas at STP is 22.4 L/mol.

So volume occupied by 0.0280 moles of `H_2O_2`
`=`0.0280 mol `xx`22.4`(L)/(mol)` `rArr` `color(blue)0.6272 L`

Appro`rArrcolor(red)627mL`