Answer to Question #131934 in General Chemistry for Sora

Question

Calculate the molality of 6.5 M aqueous solution of an acid ,HA with a density of 0.888 g cm^-3.Given the molar mass of acid is 98.0 g mol^-1

Answer 1

First we need to find the mass of this solution.

We know that $Density(p) = Mass(m)/Volume(V)$

Thus making m the subject of the formula gives

$m=pV$

(p=0.888g/cm^-3 and V=1L=1000cm³)

Thus, $m=(0.888g/cm^3) x (1000cm^3) = 888g$ (mass of the entire solution)

Secondly, we need to get the mass of water in the solution.

We have 6.5M HA, what is the mass of HA are there?

Therefore, $1 mol of HA=98.0g$

$6.5 mol of HA =?$

By cross multiplication we get;

$\frac{6.5 mol x 98.0g}{1} = 637g$ (mass of HA in the solution)

$\therefore mass of waterin the solution = 888g - 637g = 251g =0.251kg$

Calculate molality using the formula

$Molality=\frac{moles of solute}{mass in Kg of solvent}$

$= \frac{6.5 mol}{0.251Kg}$

$= 25.896 mol/Kg$