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Some Basic Concepts of Chemistry - Result Question 13 — JEE PYQ Chapterwise
19. Dissolving 120 g of urea (mol. wt. 60) in 1000 g of water gave a solution of density $1.15 \mathrm{~g} / \mathrm{mL}$. The molarity of the solution is
(2011)
(a) 1.78 M
(b) 2.00 M
(c) 2.05 M
(d) 2.22 M
Show Answer
Answer:
Correct Answer: 19. (c)
Solution:
$\begin{aligned} & \text { 19. } \text { Molarity }=\frac{\text { Moles of solute }}{\text { Volume of solution (L) }} \\ & \text { Moles of urea }=\frac{120}{60}=2 \\ & \text { Weight of solution }=\text { Weight of solvent }+ \text { Weight of solute } \\ & =1000+120=1120 \mathrm{~g} \\ & \Rightarrow \quad \text { Volume }=\frac{1120 \mathrm{~g}}{1.15 \mathrm{~g} / \mathrm{mL}} \times \frac{1}{1000 \mathrm{~mL} / \mathrm{L}}=0.973 \mathrm{~L} \\ & \Rightarrow \quad \text { Molarity }=\frac{2.000}{0.973}=2.05 \mathrm{M} \\ & \end{aligned}$





