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NEET CHEMISTRYSome Basic Concepts of ChemistryMedium

Question

6.02×10206.02 \times 10^{20} molecules of urea are present in 100 mL100 \text{ mL} of its solution. The concentration of the solution is:

A

0.01 M0.01 \text{ M}

B

0.001 M0.001 \text{ M}

C

0.1 M0.1 \text{ M}

D

0.02 M0.02 \text{ M}

Step-by-Step Solution

First, we calculate the number of moles of urea in the solution. Number of moles of urea (nn) = Given number of moleculesAvogadro’s number (NA)\frac{\text{Given number of molecules}}{\text{Avogadro's number } (N_A)} n=6.02×10206.02×1023=103 mol=0.001 moln = \frac{6.02 \times 10^{20}}{6.02 \times 10^{23}} = 10^{-3} \text{ mol} = 0.001 \text{ mol}

Next, we calculate the concentration (molarity) of the solution. Molarity (MM) is defined as the number of moles of solute dissolved in one litre of solution . Volume of solution=100 mL=0.1 L\text{Volume of solution} = 100 \text{ mL} = 0.1 \text{ L} M=Moles of soluteVolume of solution in litres=0.001 mol0.1 L=0.01 MM = \frac{\text{Moles of solute}}{\text{Volume of solution in litres}} = \frac{0.001 \text{ mol}}{0.1 \text{ L}} = 0.01 \text{ M}.

Exam Context & Concepts Covered

This question aligns with the NEET CHEMISTRY syllabus, specifically targeting concepts from Some Basic Concepts of Chemistry. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

CHEMISTRYSome Basic Concepts of Chemistrymoleculespresentsolutionconcentrationsolution

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