Ask Questions, Get Answers

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

0.22g sample of volatile compound,containing $C,H$ and $Cl$ only on combustion in $O_2$ gave 0.195g $CO_2$ and 0.0804g $H_2O$.If 0.120g of the compound occupied a volume of 37.24ml at $105^{\large\circ}$ and 768mm of pressure,calculate molecular formula of compound.


Can you answer this question?

1 Answer

0 votes
Step 1:
$\large\frac{768}{760}$$\times \large\frac{37.24}{1000}=\large\frac{0.120}{m}$$\times 0.0821\times 378$
Let formula of compound be $C_aH_6Cl_2$
$C_aH_bCl_2+(a+\large\frac{b}{4})$$O_2\rightarrow aCO_2+\large\frac{b}{2}$$H_2O+\large\frac{z}{2}$$Cl_2$
$99g\Rightarrow 44ag\;CO_2$
$6.22g\Rightarrow \large\frac{44a\times 0.22}{99}$
$\therefore \large\frac{44a\times 0.22}{99}$$=0.195$
Step 2:
Similarly $0.22g$ compound gives =$\large\frac{18\times b\times 0.22}{2\times 99}$g $H_2O$
$\large\frac{18\times b\times 0.22}{2\times 99}$$=0.0804$
Now $C_aH_bCl_2$
$12\times a+b+35.5z=99$
$\therefore C_2H_4Cl_2$
Hence (b) is the correct answer.
answered Nov 6, 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