Question

What is the mass of urea required in making 2.5 Kg of 0.25 molal aqueous solution?
1 point
56.5g
45.5g
37.5g
39.5g​

Answer 1

Answer:

• Mass of urea = 37 g.

Step-by-step explanation:

Given that,

• Moles of urea (solute) = 0.25 mole.

• Mass of water (solvent) = 1 kg = 1000 g.

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Molar mass of urea (NH₂CONH₂) =

⇒ 14 + 2 + 12 + 16 + 14 + 2

⇒ 60 g mol⁻¹.

Now, mass of 0.25 mole of urea =

⇒ 0.25 mole × 60 g mol⁻¹

⇒ 15 g.

∴ Total mass of the solution =

⇒ Mass of solute + Mass of solvent

⇒ 15 g + 1000 g

⇒ 1015 g = 1.015 kg

Therefore, 1015 kg of the solution contains urea = 15 g,

Then, 2.5 kg of the same solution will require urea =

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

$:\implies\rm{\dfrac{37.5}{1.015}\:g}$

$:\implies\bold{37\:g\:(approx.)}$

Answer 2

Given

Mass of Solution = 2.5 kg.

molality of solution = 0.25 kg.

Formula

Molality of solution =

$\frac{no \: of \: moles \: of \: solute}{weight \: of \: solvent in \: kg\: \: } </p><p>$

Molar mass of urea = 60 gram

Solution

Let the weight of solute be x kg. Subsequently , weight of the solvent will be (2.5 - x ) kg.

moles of solute =

$\frac{1000x}{60}$

Hence, molality =

$\frac{no \: of \: moles \: of \: solute}{weight \: of \: solvent \: in \: kg\: \: } \\ = > \frac{ \frac{1000x}{60} }{2.5 - x} = 0.25 \\ \\ = > \frac{1000x}{60} = 0.25(2.5 - x) \\ \\ = > \frac{1000x}{60} = 0.625 - 0.25x \\ \\ = > 1000x = 37.5 - 15x \\ \\ = > 1015x = 37. 5\\ \\ = > x = \frac{37.5}{1015} \\ \\ => x = 0.03 \: kg$

Hence, molality = 37.5 g approx .