Question

To determine the freezing point depression of a LiCl solution, Toni adds 0.411 g of LiCl to the sample test tube along with 19.7 mL of distilled water. Determine the molal concentration (m) of the resulting solution. MWLiCl

Answer 1

Answer:

LiCl = 0.492 m

Explanation:

Molal concentration is the one that indicates the moles of solute that are contained in 1kg of solvent.

Our solute is lithium chloride, LiCl.

Our solvent is distilled water.

We do not have the mass of water, but we know the volume, so we should apply density to determine mass.

Density = mass / volume

Density . volume = mass

1 g/mL . 19.7 mL = 19.7 g

We convert g to kg → 19.7 g . 1 kg / 1000g = 0.0197 kg

Let's determine the moles of LiCl

0.411 g . 1 mol / 42.394 g = 9.69×10⁻³ moles

Molal concentration (m) = 9.69×10⁻³ mol / 0.0197 kg → 0.492 m

Answer 2

The molal concentration of the resulting solution is 0.492 m. The concentration number of moles of solute in one kg of solvent.

What is Molal concentration?

It is the measure of the concentration number of moles of solute in one kg of solvent.

To calculate the molal concentration first calculated the mass of $\bold {LiCl}$,

$\rm \ mass = density \times volume$

Put the values,

m = 1 g/mL x  19.7 mL

m = 19.7 g     or

m = 0.0197 kg

Calculate the moles of LiCl:

$n =\rm \dfrac { 0.411 \ g \times 1 mol }{ 42.394 \ g }\\\\n = 9.69\times 10^{-3} \ moles$

So, now the molal concentration,

$m = \rm \dfrac { 9.69\times 10^{-3}\ mol} {0.0197 \ kg} \\\\m = 0.492$

Therefore, the molal concentration of the resulting solution is 0.492 m.

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