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Commercially available ${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$is $98\,{\text{g}}$by weight of${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$and $2\,{\text{g}}$by weight of water. Its density is ${\text{1}}{\text{.83}}\,{\text{g}}\,{\text{c}}{{\text{m}}^{ - 3}}$.Calculate the molality (m) of ${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$solution. (Take: molar mass of ${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$as${\text{98}}\,{\text{g}}\,{\text{mo}}{{\text{l}}^{ - 1}}$)
A. ${\text{500}}\,{\text{m}}$
B. ${\text{20}}\,{\text{m}}$
C. ${\text{50}}\,{\text{m}}$
D. ${\text{200}}\,{\text{m}}$

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Last updated date: 13th Jun 2024
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Answer
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Hint: We can determine the number of mole of solute (sulphuric acid) by using the mole formula. Then the amount of solvent (water) in kg. Molality is defined as the mole of solute dissolved in one kg of the solvent.
We can use the formula:
${\text{Molality}}\,{\text{ = }}\,\dfrac{{{\text{Moles}}\,{\text{of}}\,{\text{solute}}}}{{{\text{kg of solvent}}}}$
${\text{Mole}}\,{\text{ = }}\,\dfrac{{{\text{Mass}}}}{{{\text{Molar}}\,{\text{mass}}}}$

Complete Step by step answer: Determine the number of mole of sulphuric acid as follows:
${\text{Mole}}\,{\text{ = }}\,\dfrac{{{\text{Mass}}}}{{{\text{Molar}}\,{\text{mass}}}}$
The molar mass of the sulphuric acid is ${\text{98}}\,{\text{g}}\,{\text{mo}}{{\text{l}}^{ - 1}}$.
Substitute ${\text{98}}\,{\text{g}}\,{\text{mo}}{{\text{l}}^{ - 1}}$ for molar mass and ${\text{98}}$gram for mass.
$\Rightarrow {\text{Mole}}\,{\text{ = }}\,\dfrac{{98\,{\text{g}}}}{{{\text{98}}\,{\text{g}}\,{\text{mo}}{{\text{l}}^{ - 1}}}}$
${\text{Mole}}\,{\text{ = }}\,1\,{\text{mol}}$
So, the mole of sulphuric acid is$1$.
Convert the amount of solvent form g to kg as follows:
$1000\,{\text{g}}\,{\text{ = }}\,{\text{1}}\,{\text{kg}}$
$\Rightarrow 2\,{\text{g}}\,{\text{ = }}\,0.002\,{\text{kg}}$
The formula of molality is as follows:
${\text{Molality}}\,{\text{ = }}\,\dfrac{{{\text{Moles}}\,{\text{of}}\,{\text{solute}}}}{{{\text{kg of solvent}}}}$
Substitute $1$ for moles of solute (sulphuric acid) and$0.002\,{\text{kg}}$for amount of solvent (water).

${\text{Molality}}\,{\text{ = }}\,\dfrac{{\,1\,{\text{mol}}}}{{\,0.\,002\,{\text{kg}}}}$
$\Rightarrow {\text{Molality}}\,{\text{ = }}\,500\,{\text{m}}$
So, the molality of sulphuric acid solution is $500\,{\text{m}}$.

Therefore, option (A) $500\,{\text{m}}$is correct.

Note: The amount of solute is taken in the form of a mole and the amount of the solvent is taken in kg. If the molar mass of the solute is not given, it can be calculated by adding the mass of each atom of the compound. The unit of molality is mol/kg so, it is necessary to convert the unit of solvent from g to kg. The unit of molality is represented by m. The capital ‘M’ represents the unit of molarity.