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NEET CHEMISTRYMedium

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}.

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