Question

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

Answer 1

We can first find the number of grams of sulfuric acid present on 1000milliliter of the solution. Then we have to calculate the mass of the solvent and the moles of sulfuric acid. From the mass of the solvent and moles of the sulfuric acid, we have to calculate the molality of the solution.

Complete step by step solution:
Given data contains,
Volume of the solution is one liter.
Mass/volume percent of sulfuric acid is $80\%$
Density of the solution is $1.80gm{l^{ - 1}}$
We know that 80 grams of sulfuric acid is dissolved in 100 milliliter of the solution. Thus, 1000 millimeter of the solution would contain 800 grams of sulfuric acid.
Let us now calculate the mass of the solution using the density and volume of the solution.
We know that density is nothing but mass divided by volume. The formula to calculate the density is,
$Density = \dfrac{{Mass}}{{Volume}}$
We can rearrange the formula of the density to get the mass of the solution.
${\text{Mass of the solution}} = {\text{Volume}} \times {\text{Density}}$
Let us now substitute the values of volume and density to calculate the mass of the solution.
${\text{Mass of the solution}} = {\text{Volume}} \times {\text{Density}}$
Mass of the solution$= 1.80\dfrac{g}{{\not{{mL}}}} \times 1000\not{{mL}}$
Mass of the solution$= 1800g$
The mass of the solution is $1800g$.
We know that the mass of the solution is the sum of the masses of solute and solvent. We can write the formula to mass of the solution as,
${\text{Mass of solution}} = {\text{Mass of solute}} + {\text{Mass of solvent}}$
Let us now rearrange the mass of the solution to get the mass of the solvent.
${\text{Mass of solvent}} = {\text{Mass of solution}} - {\text{Mass of solute}}$
Let us now substitute the values of the mass of solution and mass of the solute.
Mass of solvent$= 1800g - 800g$
Mass of solvent$= 1000g$
The mass of the solvent is $1000g$.
Let us now calculate the number of moles of sulfuric acid.
We know that the molar mass of sulfuric acid is $98g/mol$.
Mass in moles of sulfuric acid=$\dfrac{{800g}}{{98g/mol}}$
Mass in moles of sulfuric acid=$8.16$
The number of calculated moles of sulfuric acid=$8.16moles$
We know the formula to calculate molality as,
$Molality = \dfrac{{{\text{Moles of solute}}}}{{{\text{Mass of solvent}}}}$
Let us now calculate the molality from the moles of sulfuric acid and mass of the solvent.
The number of calculated moles of sulfuric acid is $8.16moles$.
The mass of the solvent is $1000g$=$1kg$.
Molality$= \dfrac{{8.16moles}}{{1kg}}$
Molality$= 8.16moles$
The molality of the solution is $8.16m$.
Therefore,option (A) is correct.

Note:
We have to know the major benefit of using molality as a unit of concentration is that molality is dependent on the amounts of solute and solvent that is not affected by changes in temperature and pressure. In many applications, the molality carries a major advantage because the mass or the amount of a substance is frequently more essential than its volume. Another major advantage of molality is that the molality of one solute present in a solution is independent of the absence (or) presence of other solutes.