Calculate the mass of urea `(NH_(2)CONH_(2))` required in making `2.5 kg ` of `0.25` molal aqueous solution.

Explanation

0.25 molal aqueous solution to urea means that
moles of urea =0.25 mole
mass of solvent `(N_(2)CONH_(2))=60" g "mol^(-1)`
`therefore0.25` mole of urea `=0.25xx60=15g`
Mass of solution `=1000+15=1015g=1.015kg`
1.015 kg of urea solution contains 15 g urea
`therefore2.5` kg of solution contains urea `=15//1.015xx2.5=37g`