Ask Questions, Get Answers

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

Total number of atoms present in $1.0cm^3$ of solid urea (density 0.3g/$cm^3$) at $25^{\large\circ}$ are

$\begin{array}{1 1}(a)\;3.01\times 10^{21}&(b)\;2.41\times 10^{22}\\(c)\;3.01\times 10^{22}&(d)\;2.41\times 10^{23}\end{array}$

Can you answer this question?

1 Answer

0 votes
$W_{urea}=1\times 0.3g$
Also $60g$ urea has 8N atoms ($NH_2CONH_2)$
Hence (b) is the correct answer.
answered Oct 28, 2013 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