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Dissolving 120 g of urea (Mw = 60) in 1000 g of water gave a solution of density 1.15 g mL^-1. The molarity of solution is:

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2.05 M

1.78 M

2.50 M

2.22 M

answer is C.

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Detailed Solution

$\begin{array}{l} V_{sol} = \frac{(120 + 1000)}{1 .15} mL \\ M = \frac{W_{2} \times 1000}{{Mw}_{2} \times V_{sol}} = \frac{120 \times 1000}{60 \times 1120 / 1 .15} = \frac{2 \times 115}{112} = 2 .05 M \end{array}$

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