Question

Calculate the molality of $1L$ solution of 80% ${H_2}S{O_4}$ $(w/V)$ , given that the density of the solution is $1.80\,\,g\,m{L^{ - 1}}$ .
(A) $8.16$
(B) $8.6$
(C) $1.02$
(D) $10.8$

Answer 1

To solve this question, we must first understand the whole concept of molality and its calculation. Then we need to use the correct formula and correct units for the calculation of molality and then only we can have our correct answer.

Complete step-by-step solution: Molality is a measure of number of moles of solute present in $1\,\,kg$ of solvent. This contrasts with the definition of molarity which is based on a specified volume of solution.
A commonly used unit for molality in chemistry is $mol/kg$ . A solution of concentration $1\,\,mol/kg$ is also sometimes denoted as 1 molal.
Molarity (M) is one of the most common units used to quantify the concentration of a solution, representing the number of moles of solute per liter of solution.
Let’s calculate the molality of the given solution:
Step 1: In this step we will find the weight of solvent in the given solution by using its given density:
$100\,\,mL$ of solution contain $80\,g$ of ${H_2}S{O_4}$
Given, density $= \,1.8\,\,g/mL$
So, weight of given solution $= \,100 \times 1.8\,\, = \,\,180\,g$
And, therefore weight of solvent $= \,180\, - \,80\,\, = \,\,100\,g$
Step 2: And, now we will find the molality by using its formula:
Molality of a given solution $= \,\,\dfrac{{no.\,\,of\,\,moles}}{{weight\,\,of\,\,solvent\,\,in\,\,kg}}$
$= \,\,\dfrac{{(\dfrac{{80}}{{98}})}}{{(\dfrac{{100}}{{1000}})}}\,\, = \,\,\dfrac{{800}}{{98}}\,\, = \,\,8.16$
Hence, we got the molality as $8.16$.

So clearly, we can conclude that the correct answer is Option (A).

Note: Molality is the favoured concentration transmission approach because the solution’s mass of solute and solvent does not change. Practically testing in a laboratory is better. Even though due to temperature differences affecting volume in molarity and not molality, it is not as reliable as molality. Molarity error sources from volume estimation are also greater than errors from a molality balance.