Question

Calculate the boiling point of a $1M$ aqueous solution (Density - $1.04g$ $m{L}^{-1}$) of potassium chloride.
$K_b$ for water = $0.52Kkgmo{l}^{-1}$.
Atomic mass of $K=39g/mol$
Atomic mass of $Cl=35.5g/mol$

• A. a. $374.07K$
• B. b. $274.11K$
• C. c. $378.11K$
• D. d. $473.11K$

Answer 1

The correct option is A a. $374.07K$

$\text{Molar mass of}KCl=74.5g/mol$
$\text{Molarity of KCl}=1M$
$\text{Mass of solution}=V\times d=1000mL\times 1.04g/mL=1040g$
$\text{Mass of solvent = Mass of solution - Mass of solute}$
$\text{Mass of solvent}=1040-74.5=965.5g$

$\text{Molality}=\frac{\text{Number of moles of solute}\times 1000}{\text{Weight of solvent in g}}$

$m=\frac{1\times 1000}{965.5}=1.035m$

$KCl$ is a strong electrolyte
$KCl\to K^{+}+C{l}^{-}$
$i=2$

$\Delta T_b=i\times K_b\times m$

$=2\times 0.52\times 1.035$

$=1.077K$

$\text{Boiling point of solution}=T_0+\Delta T_b$

$=373+1.077$

$=374.077K$