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Question: What is the mass of urea required for making \(0.25\)molal aqueous solution...

What is the mass of urea required for making 0.250.25molal aqueous solution

A

37 g

B

25 g

C

125 g

D

27.5 g

Answer

37 g

Explanation

Solution

No. of moles of urea = 0.25 moles

Mass of solvent = 1 kg = 1000 g

Molar mass of urea = 60 mol−160 \mathrm {~mol} ^ { - 1 }

Mass of urea in 1000 g of H2O=0.25×60=15 g\mathrm { H } _ { 2 } \mathrm { O } = 0.25 \times 60 = 15 \mathrm {~g}

Total mass of solution =1000+15=1015 g= 1000 + 15 = 1015 \mathrm {~g}

1.0151.015 kg solution contains 15 g urea

2.5 kgsolution will contain =151.015×2.5=37 g= \frac { 15 } { 1.015 } \times 2.5 = 37 \mathrm {~g}