logo

Ask Questions, Get Answers

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

A mixture of $HCOOH$ and $H_2C_2O_4$ is heated with Conc.$H_2SO_4$.The gas produced is collected and on treating with KOH solution the volume of the gas decreases by $\large\frac{1}{6}$th.Calculate molar ratio of two acids in original mixture.

$(a)\;2 : 1\qquad(b)\;4 : 1\qquad(c)\;3 : 2\qquad(d)\;1 : 2$

Can you answer this question?
 
 

1 Answer

0 votes
Step 1:
Let a mole of HCOOH and b mole of $H_2C_2O_4$ are present in original mixture.
Mole of $CO$ formed =a+b
$a\Rightarrow$ From HCOOH
$b\Rightarrow$ From $H_2C_2O_4$
Mole of $CO_2$ formed =b
Total mole of gases =$a+b+b=a+2b$
Step 2:
$\therefore CO_2$ is observed by KOH and volume reduces by $\large\frac{1}{6}$
Mole of $CO_2=\large\frac{1}{6}$$(a+2b)$
$b=\large\frac{1}{6}$$(a+2b)$
$\large\frac{a}{5}$$=4$
$a : b=4 : 1$
Hence (b) is the correct option.
answered Oct 31, 2013 by sreemathi.v
 

Related questions

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