(K \_{ b }) for water is (0.52, K / m). Then (0.1 ,m) soluti... | Chemistry | SolveeIt

Question

$K _{ b }$ for water is $0.52\, K / m$ . Then $0.1 \,m$ solution of $NaCl$ will boil approximately at:

$100.52^{\circ} C$

$100.052^{\circ}C$

$101.04^{\circ}C$

$100.104^{\circ}C$

Answer

$100.104^{\circ}C$

Explanation

Solution

Given
$0,1\,m \,Nacl$
$Kb$ for water $=0.52\, km$
Boiling point of solution $=?$
solution
$\Delta T_{b}=k_{b} \times m \times i$
$|$ for $Nacl \,\,i=2 |$
$\Delta T_{b}=2 \times 0.1 \times 0.52$
$\Delta T_{b}=0.104$
Boiling point of solution
$=100+\Delta T_{b}$
$=100+0.104=100 . 104$

Chemistry Question on Colligative Properties

Solution