Question

Hello friends☺️☺️✌️❤️
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Calculate the molality of a Sulphuric Acid solution in which the mole fraction of water is 0.85 .
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Answer 1

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Answers :

The relationship between mole fraction and molality is given as

m=1000 X xB/(1-XB)X Molecular weoght of solvent

The tricky part in this question is observing which is solute and solvent.

Mole fraction of water is 0.85 which means it is not the solute but it is the SOLVENT.

We know that mole fraction of solute+mole fraction of solvent=1

Therefore, xB=0.15

Substitute it in given equation,

1000(0.15)/(0.85)(18)

Which is equal to 9.8.

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by @neel

Answer 2

SOLUTION ☺️❣️

We have mole fraction of solute + solvent

= 1

Here, Mole fraction of water is 0.85 so mole fraction of sulphuric acid will be (1-0.85)= 0.15

Hence, each mole of solution 0.15 mole of sulphuric acid is dissolved

So, mass of solvent (water) is

For 1 mole of water = 18gm

So, 0.85 mole of H2O = 18×0.85= 15.3gm

=) 0.0153 kg

Thus,

=) molality (m)= no. of moles of solute/ mass of solvent

=) Hence, molality of the solution (m)

=) 0.15/0.0153 = 9.8 m

(nearly = 10)

HOPE it helps ✔️