Question

The density of $ 2.43M $ aqueous methanol ( $ C{H_3}OH $ ) is $ 0.976g/mL $ . What is the molality of the solution (Molecular weight of $ C{H_3}OH = 32 $ )?
(a) $ 27.3m $
(b) $ 0.273m $
(c) $ 7.23m $
(d) $ 2.73m $

Answer 1

Hint: For these questions, first we have to know what is the molarity and molality of a given solution. Then how to find the molality of the solution when molecular mass, the molarity and the density of the solution is given.
Formula used
Molarity:
 $ M = \dfrac{n}{V} $
Where, $ M $ is the molarity of the solution, $ n $ is the number of moles present in the solution and $ V $ is the volume of the solution in the litres.
Density:
 $ d = \dfrac{W}{V} $
Where, $ d $ is the density of the solution, $ V $ is the volume in the litres and $ W $ is the weight is kilograms.
Molality:
 $ m = \dfrac{n}{W} $
Where, $ m $ is the molality of the solution, $ n $ is the number of moles present in the solution and $ W $ is the weight of solvent in kilograms.

Complete step-by-step answer
We will start with the definition of the molarity:
Molarity: The number of moles of the solute present in per one litre of the solution is called molarity.
Now, moving towards the definition of molality:
Molality: Molality is defined as the number of moles of solute present in per one kilogram of the solvent.
Given data in the question:
Density of the solution, $ d = 0.976g/mL $
Molarity of the solution, $ M = 2.45M $
Molecular weight of $ C{H_3}OH $ , $ mol = 32 $
For finding the molality, we need the number of moles of solute and the mass of the solvent.
The molarity is given, which states that $ 2.45M $ moles are present in the one litre of solution,
The mass of the solute, $ W = n \times mol = 2.45 \times 32 = 78.4g $
Now, finding the mass of solvent, for this we have to find the mass of the solution, we can find out the mass of solution using density,
 $ d = \dfrac{W}{V} $
 $ \Rightarrow W = V \times d $
Putting the values of volume and density,
 $ \Rightarrow W = V \times d = 0.976 \times 1000 = 976g $
Hence, the mass of the solution is $ 976g $
Mass of solvent= mass of given solution-mass of calculated solute
 $ \Rightarrow $ Mass of solvent $ = (976 - 78.4)g = 897.6g $
Hence, the mass of the solvent is $ 897.6g $
Hence, now when we know the number of moles and mass of the solvent we can find out the molality of the solution.
Now, putting the values of the moles and mass of solvent,
 $ m = \dfrac{n}{W} = \dfrac{{2.45 \times 1000}}{{897.6}} \approx 2.73m $
So, the molality of the solution is $ 2.73m $ .
Hence, the correct option is (d) $ 2.73m $ .

Note
While measuring the more accurate concentration, we prefer for the molality rather than molarity, because we take volume of the solution in molarity, the volume may change due to the temperature but on the weight it does not affect.