$6.02 \times 10^{20}$ molecules of urea are present in $100 \mathrm{~mL}$ of its solution. The concentration…

$6.02 \times 10^{20}$ molecules of urea are present in $100 \mathrm{~mL}$ of its solution. The concentration of solution is
  1. $0.02 \mathrm{M}$
  2. $0.01 \mathrm{M}$
  3. $0.001 \mathrm{M}$
  4. $0.1 \mathrm{M}$

Solution

Given, number of molecules of urea $=6.02 \times 10^{20}$
$\begin{aligned}
& \therefore \text { Number of moles }=\frac{6.02 \times 10^{20}}{N_A} \\
& =\frac{6.02 \times 10^{20}}{6.02 \times 10^{23}}=1 \times 10^{-3} \mathrm{~mol}
\end{aligned}$
Volume of the solution
$=100 \mathrm{~mL}=\frac{100}{1000} \mathrm{~L}=0.1 \mathrm{~L}$
Concentration of urea solution
$\begin{aligned}
\left(\text { in } \mathrm{mol} \mathrm{L}^{-1}\right) & =\frac{1 \times 10^{-3}}{0.1} \mathrm{mol} \mathrm{L}^{-1} \\
& =1 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1}=0.01 \mathrm{~mol} \mathrm{~L}^{-1}
\end{aligned}$

Asked in: NEET 2013 (All India)

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