Find the mass of urea (CH4N2O) needed to prepare 48.5 g of a solution in water in which the mole fraction of urea is 7.60×10−2.

13.62 g of urea

Explanation:

mass of water = 48.5 g

number of moles of water = 48.5 g/18g/mol = 2.7 moles of water

mole fraction of urea = 7.60×10−2

but mole fraction = mole of substance/ total number of moles

Let the number of moles of urea be x

7.60×10−2 =x/2.7 + x

7.60×10−2(2.7 +x) = x

0.21 + 0.076x = x

0.21 = x-0.076 x

0.21 = 0.924x

x = 0.21/0.924

x = 0.227 moles of urea

mass of urea in solution = number of mole * molar mass

= 0.227 moles * 60 g/mol = 13.62 g of urea

2b2t

explanation:

2b2t