Question

Density of 2.03 M aqueous solution of acetic acid is $1.017gm{L^{ - 1}}$ molecular mass of acetic is 60. Calculate the molality of the solution.
$A.$ 2.27
$B.$ 1.27
$C.$ 3.27
$D.$ 4.27

Answer 1

Hint: Molality is the measure of moles of solute per kg of solvent. If volume and density are given, we can find how many grams of solute we have. From grams we can find moles. Once we have moles, we can divide by the kg of solvent to get molality. The specific or easiest steps you take depend on what you are given.

Complete answer:
Here we have, Molarity of solution = $2.03M$ and density of solution = $1.017gm{L^{ - 1}}$
​Also we have molar mass of acetic acid = $60gmo{l^{ - 1}}$
We know, Molality = $\dfrac{{No.\;of\;Moles}}{{Mass\;of\;solvent(kg)}}$
We have the number of moles but we need to find the mass of solvent. As we know molarity is $\dfrac{{No.\;of\;moles}}{{volume\;of\;solution\;(in litres)}}$
​Molarity equals the number of moles if we have 1 L solution. Therefore, Number of moles = 2.03 in 1 L or 1000 mL solution
$Density = \dfrac{{Mass}}{{Volume}}$or we can also say, $Mass = Density \times Volume$
Now, Mass = 1.017​1000= 1017 g
Therefore Mass of solution = 1017 g
Now, Mass of acetic acid in solution = Number of moles of $C{H_3}COOH$$ \times $ Molar mass= 2.03$ \times $ 60 = 121.8 g
Therefore, Mass of solvent = mass of solution - mass of acetic acid $(C{H_3}COOH)$
= 1017-121.8
= 895.2 g or 895.2$ \times {10^{ - 3}}$ kg
$Molality = \dfrac{{Number\;of\;moles}}{{Mass\;of\;Solvent(kg)}}$
= $\dfrac{{2.03}}{{895.2 \times {{10}^{ - 3}}}}$ kg
= 2.27 m

Hence our answer is A, 2.27 m is the molality of the solution.

Note: Molality is also called molal concentration and is defined as the amount of substance of solute, divided by the mass of the solvent m. In other words molarity is the number of moles of solute per litre of solution. For converting it to density we multiply by the number of moles by the molecular mass of the compound.