Question

Dissolving 120 g of urea (mol. wt. 60) in 1000 g of water gave a solution of density 1.15 g/mL. The molarity of the solution is

• A. 1.78 M
• B. 2.00 M
• C. 2.05 M
• D. 2.22 M

Answer 1

Answer: (C)

2.05 M

Answer 2

$molarity = \frac{Moles \, of \, solute }{Volume \, of \, solution \, (L)}$
Moles of urea $= \frac{ 120}{60} = 2$
Weight of solution = Weight of solvent + Weight of solute
= 1000 + 120 = 1120 g
$\Rightarrow \, \, \, \, \, \, \, Volume = \frac{ 1120 g }{ 1.15g/mL } \times \frac{1}{ 1000 mL/L} = 0.973 L$
$\Rightarrow \, \, \, \, \, Molarity = \frac{2.000}{0.973} = 2.05 M$