$6.02 imes 10^{20}$ molecules of urea are present in $100 ext{ mL}$ of its solution. The concentration of

Question

$6.02 	imes 10^{20}$ molecules of urea are present in $100 	ext{ mL}$ of its solution. The concentration of the solution is:

Accepted Answer

First, we calculate the number of moles of urea in the solution.
Number of moles of urea ($n$) = $\frac{	ext{Given number of molecules}}{	ext{Avogadro's number } (N_A)}$ 
$n = \frac{6.02 	imes 10^{20}}{6.02 	imes 10^{23}} = 10^{-3} 	ext{ mol} = 0.001 	ext{ mol}$

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

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