Question

Calculate the mass of urea required in making 2.5 kg of 0.25 molal aquas solution

Answer 1

Answer:

given: molality=0.25 and W in Kg =2.5 and molar mass of urea is 60

molality =no of moles/weight of solution in kg

substituting the value we get

no. of moles = 0.625

now

we know n = given mass/molar mass

0.625x60 is mass required

final ans is 37.5 g required

Answer 2

Answer:

• Mass of urea required = 37 g.

Step-by-step explanation:

Given that,

• Moles of urea = 0.25 mole.
• Mass of water (solvent) = 1 kg = 1000 g.

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Molar mass of urea = 14 + 2 + 12 + 16 + 14 + 2 = 60 g/mol.

∴ Mass of 0.25 moles of urea = 0.25 mole × 60 g/mol = 15 g.

(∵ No. of moles = mass/molar mass)

Now, total mass of the solution = 1000 + 15 g = 1015 g = 1.015 kg.

Thus, 15 g of urea is present in 1.015 kg of the solution.

∴ Mass of urea required for 2.5 kg of solution,

$\implies\rm{\dfrac{15\:g}{1.015\:kg}\times{}2.5\:kg}$

$\implies\rm{37\:g.}$