Question

What is the boiling point of 1 molal aqueous solution of NaCl [kb=0.52 K molal -1
]

Answer 1

The boiling point of 1 molal aqueous solution of NaCl is 101.04 °C.

Explanation:

Solutions that do not obey Raoult's law are called non-ideal solutions. The features of a non-ideal solution that are dependent on the number of solute particles dissolved in a solvent are known as colligative properties.

The formula is used to calculate the boiling point of NaCl solution:

Δ$T_{b} = imK_{b}$

where;

Δ$T_{b} = change\ in\ temperature$

$i = Van't\ Hoff\ factor$

$m = molality$

$K_{b} = molal\ boiling\ point\ constant$

Δ $T_{b} = T_{f} - T_{i}$

$T_{f} = Final\ Boiling\Temperature\ of\ solution\\T_{i} = Boiling\Temperature\ of\ pure\ solvent$

Given:

$m = 1m$

$K_{b} = 0.52 m^{-1}$ °C

$T_{i} = 100$ °C

$i = 2$

$T_{f} - T_{i} = imK_{b}$

$T_{f} - 100 = 1 * 0.52 * 2\\T_{f} - 100 = 1.04\\T_{f} = 1.04 + 100\\T_{f} = 101.04\ degree\ Celsius$