Ask Questions, Get Answers

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

A solid mixture 5g consists of lead nitrate and sodium nitrate was heated below $600^{\large\circ}C$ until weight of residue was conotaut.If the loss in weight is 23%,find the amount of lead nitrate and sodium nitrate mixture.

$\begin{array}{1 1}(a)\;3.32g,1.68g&(b)\;2.38g,1.28g\\(c)\;1.14g,1g&(d)\;3.2g,2g\end{array}$

question is wrong it should be 28% as per to the answer
Can you answer this question?

1 Answer

0 votes
Step 1:
$Pb(NO_3)_2\rightarrow PbO+2NO_2+\large\frac{1}{2}$$O_2$
$NaNO_3\rightarrow NaNO_2+\large\frac{1}{2}$$O_2$
The loss in weight for 5g mixture $=5\times \large\frac{28}{100}$$=1.4g$
Residue left =$5-1.4g=3.6g$
Step 2:
The residue contains $PbO+NaNO_2$
331g $pb(NO_3)_2$ gives=223g PbO
$a$g $Pb(NO_3)_2$ gives =$\large\frac{223\times a}{331}$g pbo
Similarly 85g $NaNO_3$ gives =$69g$ $NaNO_2$
$b$g $NaNO_3$ gives =$\large\frac{69\times b}{85}$ g $NaNO_2$
$\large\frac{223\times 9}{331}+\large\frac{69}{85}$$\times b=3.6$-----(2)
Step 3:
From (1) & (2) we get
Hence (a) 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