Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

$6.0\times 10^{20}$ molecules of urea are present in 100ml of its solution .The concentration of urea solution is

$\begin{array}{1 1}(a)\;0.01M&(b)\;0.001M\\(c)\;0.1M& (d)\;0.02M\end{array} $

Can you answer this question?

1 Answer

0 votes
$6.023\times 10^{23}$ molecule of urea=1mol
$6.0\times 10^{20}$ molecule of urea =$\large\frac{6.0\times 10^{20}}{6.023\times 10^{23}}$$=10^{-3}$ mol
$\therefore 10^{-3}$ mol of urea are present in 100 ml
$\therefore$ moles of urea present in 1000ml=$10^{-3}\times \large\frac{1000}{100}$$=10^{-2}$
$\therefore$ molarity of solution =$10^{-2}M=0.01$M
Hence (a) is the correct option.
answered Mar 25, 2014 by sreemathi.v

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App