Ask Questions, Get Answers

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

$8.0575\times 10^{-2}kg$ of Glauber's salt is dissolved in water.Obtain $1clm^3$ of a solution of density $1077.2kgm^{-3}$.Calculate the molarity and the mole fraction of $Na_2SO_4$ in solution .

$(a)\;0.2502m,0.24m,4.3\times 10^{-3}\\(b)\;0.3m,0.1m,2\times 10^{-2}\\(c)\;.15m,0.3m,1.3\times 10^{-3}\\(d)\;.3m,0.156m,1.3\times 10^{-2}$

Can you answer this question?

1 Answer

0 votes
Step 1:
Mole weight $Na_2SO_4.10H_2O=322$
Weight of $Na_2SO_4$ in $8.0575\times 10^{-2}kg$ glauber salt
$\Rightarrow \large\frac{142\times 8.0575\times 10^{-2}}{322}$
$\Rightarrow 3.5533\times 10^{-2}kg$
M of $Na_2SO_4=\large\frac{3.5533\times 10^{-2}}{142\times 10^{-3}\times 1}$
$\Rightarrow 0.2502m$
Step 2:
Molality of $Na_2SO_4=\large\frac{3.5533\times 10^{-2}}{142\times 10^{-3}\times 1}$
$\Rightarrow \large\frac{3.5533\times 10^{-2}}{142\times 10^{-3}\times \Large\frac{1041.667}{10^{-3}}}$
$\Rightarrow 0.24m$
Step 3:
Mole fraction of $Na_2SO_4=\large\frac{Mole\;of\;Na_2SO_4}{Mole\;of\;Na_2SO_4+Mole\;of\;H_2O}$
$\Rightarrow \large\frac{3.5533\times 10^{-2}/142\times 10^{-3}}{3.5533\times 10^{-2}/142\times 10^{-3}+1041.667/18}$
$\Rightarrow 4.3\times 10^{-3}$
Hence (a) is the correct answer.
answered Nov 6, 2013 by sreemathi.v
Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App