Search any question & find its solution
Question:
Answered & Verified by Expert
$6.02 \times 10^{20}$ molecules of urea are present in $100 \mathrm{~mL}$ of its solution. The concentration of solution is
Options:
Solution:
1078 Upvotes
Verified Answer
The correct answer is:
$0.01 \mathrm{M}$
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
$\begin{aligned}
\left(\text { in } \mathrm{mol} \mathrm{L}^{-1}\right) & =\frac{1 \times 10^{-3}}{0.1} \mathrm{mol} \mathrm{L}^{-1} \\
& =1 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1}=0.01 \mathrm{~mol} \mathrm{~L}^{-1}
\end{aligned}$
$\begin{aligned}
& \therefore \text { Number of moles }=\frac{6.02 \times 10^{20}}{N_A} \\
& =\frac{6.02 \times 10^{20}}{6.02 \times 10^{23}}=1 \times 10^{-3} \mathrm{~mol}
\end{aligned}$
Volume of the solution
$=100 \mathrm{~mL}=\frac{100}{1000} \mathrm{~L}=0.1 \mathrm{~L}$
Concentration of urea solution
$\begin{aligned}
\left(\text { in } \mathrm{mol} \mathrm{L}^{-1}\right) & =\frac{1 \times 10^{-3}}{0.1} \mathrm{mol} \mathrm{L}^{-1} \\
& =1 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1}=0.01 \mathrm{~mol} \mathrm{~L}^{-1}
\end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.