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

3.37 m

Step-by-step 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 = (Mass\ taken)/(Molar\ mass)`

Thus,

`Moles= (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