logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  JEEMAIN and AIPMT  >>  Chemistry  >>  Solutions
0 votes

The boiling point of water at $735$Torr is $99.07^{\circ}C$ . The mass of $NaCl$ added in $100\;g$ water $(K_b=0.51\;K\;kg\;mol^{-1})$ to make its boiling point $100^{\circ}C$ is

$\begin{array}{1 1} 10.68\;g \\ 5.34 \;g\\ 2.67\;g \\ 26.7\;g \end{array} $

Can you answer this question?
 
 

1 Answer

0 votes
Answer :$ 5.34 \;g$
$m = \large\frac{T_b}{K_b}$
$\quad= \large\frac{(100-99.07)}{0.51}$$mol\;kg^{-1}$
Molality of $NaCl=\large\frac{1}{2} \bigg( \large\frac{100 -99.07}{0.51} $$ mol\;kg^{-1}\bigg)$
Mass of NaCl in 100 g water $= \large\frac{1}{2} \bigg( \large\frac{1}{10} \bigg)\bigg( \frac{0.93}{0.51}\bigg)$$ (58.5)g=5.33\;g$
answered Aug 6, 2014 by meena.p
 

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
...