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Q.
$6.02 \times 10^{20}$ molecules of urea are present in 200 mL of its solution. The concentration of urea solution is $(N_{0} = 6.02\times 10^{23}\,mol^{-1})$
Solutions
Solution:
Molarity $(mol\,L^{-1}) =\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$
$6.02 \times 10^{23}$ molecules of urea = 1 mole
$\therefore 6.02 \times 10^{20}$ molecules of urea
$=\frac{6.02\times 10^{20}}{6.02\times 10^{23}}=1\times 10^{-3}\,mol$
$200\,ml =0.20\,L$
Molarity $=\frac{1\times 10^{-3}}{0.20}=0.02\,M$