Question

What is the molality of the resulting solution when 15.0 g of Licl is combined with 105 g of water? 0/1 point 0 1.42 m 2.95 m 3.37 m 1.25 m 0 0 Incorrect

Answer 1

Answer:

3.37 m

Explanation:

Number of moles of solute present in 1 kg of solvent is termed as molality

It is represented by 'm'.

Thus,

$Molality\ (m)=\frac {Moles\ of\ the\ solute}{Mass\ of\ the\ solvent\ (kg)}$

Given that:

The mass of LiCl = 15.0 g

Molar mass of LiCl = 42.394 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{15.0\ g}{42.394\ g/mol}$

$Moles= 0.3538\ mol$

Mass of the solvent = 105 g

Also, 1 g = 0.001 g

So,

Mass of water (solvent) = 0.105 kg

Molality is:

$Molality\ (m)=\frac {0.3538}{0.105}\ m$

Molality = 3.37 m