Question

A solution containing 3.1g of bacl2 in 250 g of water boils at 100.083 c. Calculate the values of van't hoof factor and molality of cacl2 in the solution

Answer 1

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here is your answer

Answer 2

Concept:

The term "total moles of a solute contained in a kilogram of a solvent" is used to describe molality. Molal concentration is another name for molality. It is a measurement of a solution's solute concentration.

Given:

Calculate the values of the van't Hoff factor and molality of $BaCl_{2}$ in the solution.

Find:

Mass of $BaCl_{2}$ = 3.1 g

Mass of water = 250 g

Tempeture = 100.083 °C

Solution:

Mass of  = 3.10 g

Mass of water = 250 g = 0.250 kg

The molality of the solution can be calculated as:

$m = \frac{number of moles of solute}{Mass of solvent in Kg }$ = $\frac{3.10 g}{208.3 g/mol * 0.250 Kg }$ = 0.595 mol/Kg

The boiling point of the solution = $T_{b}$ = 100.083 °C = 373.083 K

Boiling point of water = T = 100 °C = 373 K

ΔT = $T_{b} - t = 373.083 K - 373 K = 0.083 K$

Δ$T_{b}$ = $ik_{f}$ * m

0.083 K = i * 0.52 K kg/mol * 0.0595 mol/kg

i =  0.083 / 0.03094

i = 2.68

Hence, the value of molality of the $BaCl_{2}$ solution is 0.0595 mol/kg.

The van't Hoff factor in this solution is 2.68.

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