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# Calculate the molality of a sulphuric acid solution of specific gravity $1.2$ containing $27\% {H_2}S{O_4}$ by weight.

Last updated date: 12th Sep 2024
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:Hint:To calculate the molality of the sulphuric acid solution we need the number of moles of solute and the weight of the solvent. So with the help of given quantities, we will calculate the number of moles of solute and weight of solvent to calculate molality.
Formula Used:
Number of moles
$n = \dfrac{W}{M}$
Where $n$ represents the number of moles,
$W$ represents the weight of a substance,
$M$ represents the molecular weight of the substance,
Molality of Solution
$m = \dfrac{n}{w}$
Where $m$ represents the molality of solution,
$n$ represents the number of moles of solute,
$w$ represents the weight of the solvent.

To solve this numerical based on molality we will consider that the weight of the total solution present is $1000g$. So we considered the weight of the solution to solve the numerical easily.
Now we will understand the meaning of a solution containing $27\% {H_2}S{O_4}$ by weight. It simply means that $270g$ of ${H_2}S{O_4}$ is present in the total solution. So now the weight solution as $1000g$ and weight of solute $270g$ we can easily calculate the weight of the solvent. So the weight of the solvent can be calculated as
$w = {w_{solution}} - {w_{solute}}$Substituting the values we get,
$\Rightarrow w = 1000 - 270$
$\Rightarrow w = 730g$
Now we calculated the weight solution and now we need the number of moles of solute to calculate molality. We know that the number of moles is given by $n = \dfrac{W}{M}$. We have $W = 270g,M = 2 \times 1 + 32 + 16 \times 4 = 98g/mol$ to substitute the values we get,
$n = \dfrac{W}{M} = \dfrac{{270}}{{98}}$
$\Rightarrow n = 2.755$Moles
Now we have calculated the number of moles of solute $n = 2.755$ and weight of solvent $w = 730g$ we can easily calculate the molality by substituting the values in the formula,
$m = \dfrac{n}{w}$
$\Rightarrow m = \dfrac{{2.755}}{{730}} \times 1000$ We multiplied by $1000$ to convert weight in kilograms
$\Rightarrow m = 3.77$

Therefore, the molality of the sulphuric acid solution is $3.77m$.
Note:
Molality is defined as no. of moles of solute present in 1kg of solvent while Molarity is defined as the no. of moles of solute present in 1L of solution. Molality is temperature independent while molarity is temperature dependent.