Question

Calculate the mass of urea co(nh2)2 present in 0.25 molal aqueous solution of urea

Answer 1

Given that

Mass of solution = 2.5 Kg

Molality = 0.25 m

Molar mass of urea (NH2CONH2) =14 + 2 × 1 + 12 + 16 + 14 + 2 × 1 = 60 g mol−1

Let mass of urea (NH2CONH2) = w gram

Number of moles of urea = w / 60 = 0.01667 mol

Mass of urea in Kg = w/1000 = 0.001w kg

Mass of solvent = total mass - mass of solute = 2.5 - 0.001 w kg

Use the above formula we get

Cross multiply and solve for the value of w we get

w = 37 g of urea

Hope this helps you ☺️☺️⭐✨✨⭐✌️✌️

Answer 2

_/\_Hello mate__here is your answer--

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Molar mass of urea (NH2CONH2)

= 2(1 × 14 + 2 × 1) + 1 × 12 + 1 × 16

= 60 g mol^−1

0.25 molar aqueous solution of urea means:

1000 g of water contains 0.25 mol

= (0.25 × 60)g of urea

= 15 g of urea

That is,

(1000 + 15) g of solution contains 15 g of urea

Therefore, 2.5 kg (2500 g) of solution contains

= (15× 2500)/( 1000+ 15 )

=36.95 g

= 37 g of urea (approximately)

Hence, mass of urea required = 37 g

I hope, this will help you.☺

Thank you______❤

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