The molality of a urea solution in which ` 0.0200 g ` of urea, ` (NH_(2))CONH_(2) ` is added to ` 0.400dm_(3) `of water at STP is

Options

A. `0.555`molal
B. `5.55xx10^(-4)`molal
C. `8.33xx10^(-2)`molal
D. `33.3`molal

Answer

C

Explanation

`1 dm^(3)=1000cm^(3)=1L=1kg`
`0.400dm^(3)of H_(2)O` at `STP=0.400kg` of `H_(2)O`
Moles or urea `=(0.0100)/(60)mol(0.0200)/(60)`
`:.` Molality `=("Moles of solute(urea)")/("kg of solvent")`
`=(0.010//60 mol)/(0.300kg)(0.020//60)/(0.400)`
`= 8.33xx10^(-4) mol kg^(-)`