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Question: The molality of the \(90\% \) \((mass.volum{e^{ - 1}})\)\[{H_2}S{O_4}\] solution is [ Given: densi...

The molality of the 90%90\% (mass.volum{e^{ - 1}})$$${H_2}S{O_4}$$ solution is [ Given: density of solution = 1.8g.m{l^{ - 1}}]A)] A)1.8B) B)48.4C) C)10.20D) D)94.6$

Explanation

Solution

Sulfuric acid, also known as oil of vitriol, is a mineral acid with the molecular formula H2SO4{H_2}S{O_4} and is made up of the elements sulphur, oxygen, and hydrogen. It is a colourless, odourless, viscous liquid that is miscible in all concentrations of water. Since it is an oxidant and has a strong acidic disposition, the acid in its pure form is extremely corrosive to other products.

Complete answer:
Molality is a property of solutions that is determined by dividing the number of moles of a solute by the number of kilogrammes of the solvent. Molality, unlike molarity, which is determined by the volume of the solution, is determined solely by the mass of the solvent. Since volume is affected by temperature and pressure, molarity is affected as well. Since mass does not differ with atmospheric conditions, using weight may be advantageous in certain situations. Molality is used when dealing with a wide variety of temperatures, for example.
If we know the mass of the solute and solvent in a solution, calculating molality is easy. Molality is a property that is independent of the quantity being calculated since it is an intensive property. This holds true for all homogeneous solution concentrations, whether we're looking at a 1.0L1.0L or a 10.0L10.0L sample of the same solution. The molality, or concentration, remains unchanged.
According to the question,
The means of 90%90\% mass.volume−1mass.volum{e^{ - 1}} H2SO4{H_2}S{O_4} is that 90g90g H2SO4{H_2}S{O_4} is present in 100ml100mlof solution.
It is given that the density of H2SO4{H_2}S{O_4} solution is 1.8g.ml−11.8g.m{l^{ - 1}}.
It means that 90g90g H2SO4{H_2}S{O_4} is present in (1.8×100)(1.8 \times 100) solution.
The mass of water can be calculated as: 180−90=90g180 - 90 = 90g
H2SO4{H_2}S{O_4} in 90g90gwater –
∴\therefore Molality =No.  of  moles  of  soluteMass  of  solvent  (Kg) = \dfrac{{No.\;of\;moles\;of\;solute}}{{Mass\;of\;solvent\;(Kg)}}
=90×100098×90= \dfrac{{90 \times 1000}}{{98 \times 90}}
=10.204mol.kg−1= 10.204mol.k{g^{ - 1}}
⇒Molality=10.204m\Rightarrow Molality = 10.204m
The correct option is C.

Note:
Since the volume of a solution is affected by the ambient temperature and pressure, mass is a better way to measure it. Molality (rather than molarity) is the appropriate measurement in these situations.