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Q.
What is the mass of urea required for making $2.5\, kg$ of $0.25$ molal aqueous solution?
Solutions
Solution:
No. of moles of urea $= 0.25$ moles
Mass of solvent $= 1\, kg = 1000\, g$
Molar mass of urea $= 60\, g \,mol^{-1}$
Mass of urea in $1000\, g$ of $H_2O = 0.25 \times 60 = 15\, g$
Total mass of solution $= 1000 + 15 = 1015 \,g$
$1.015 \,kg$ solution contains $15\, g$ urea
$2.5\, kg$ solution will contain $= \frac{15}{1.015} \times 2.5 = 37\,g$