Chemistry, asked by prashant0005, 6 months ago

You are supplied with 25% (by mass) aqueous solution of Sulphuric acid(98g/mol) having a density of 1.84g/cc. (3a)Calculate the molality of the solution. (show calculation)​

Answers

Answered by GuriSingh07
0

Answer:

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Answered by rsagnik437
4

Given:-

→ Concentration [by mass] = 25%

Molar mass of Sulphuric acid = 98g/mol

Density of the solution = 1.84g/cc

To find:-

→ Molality of the solution.

Solution:-

It is clear that sulphuric acid is solute and water is solvent.

Let the total mass of the solution be 100g.

Then mass of solute [sulphuric acid] :-

= 100/100 × 25

= 25g

And mass of solvent [water] :-

= 100 - 25

= 75g

Number of mole in 25g of solute :-

= Given Mass/Molar mass

= 20/98

= 0.255 mole [approximately]

Now, let's convert the mass of solvent from 'grams' to 'kilograms' :-

=> 1g = 0.001kg

=> 75g = 75(0.001)

=> 0.075kg

Molality of a solution :-

= Moles of solute/Mass of solvent in kg

= 0.255/0.075

= 255/75

= 3.4m

Thus, molality of the solution is 3.4m .

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