Question

Find the mass of urea (CH₄N₂O) needed to prepare 65.0 g of a solution in water in which the mole fraction of urea is 0.0981.

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Answer 1

Let, ‘U’ be the moles of urea and ‘W’ be the moles of water. Then, mole fraction of urea can be written as follows:

Mole fraction of urea = U/U+W

0.0981 = U/U+W

0.0981(U+W) = U

0.0981U + 0.0981W = U

0.0981W = U – 0.0981U

W =(1-0.0981)U / 0.0981

W = 9.194 U    -----------------------(1)

Molar mass of urea is 60g/mol and that of water is 18g/mol

Since, weight of solution is 65; Total of weight of urea and water should be 65.

Weight of urea + weight of water = 65

(Molar mass of urea x moles of urea ) + (molar mass of water x moles of water) = 65

(60 x U) + (18 x W) = 65

Substitute value of W from equation (1)

(60 x U) + (18 x 9.194 U) = 65

60 x U + 165.5 x U = 65

225.5 x U = 65

U = 65 / 225.5

U = 0.288 mol

Thus, moles of urea used is 0.228 moles. Weight of 0.288 moles of urea can be calculated as follows:

Weight of urea = moles of urea x molar mass of urea

Weight of urea = 0.288 mol x 60 g/mol

Weight of urea = 17.3 g

Thus, mass of urea needed to prepare 65.0g of solution is water is 17.3g