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  • Calculate the molality of a 1-liter solution of 93%   (weight/volume). The dens

Calculate the molality of a 1-liter solution of 93%  H_{2}SO_{4} (weight/volume). The density of the solution is 1.84 g /ml

Option: 1

10.43


Option: 2

20.36


Option: 3

12.05


Option: 4

14.05


Answers (1)

best_answer

Given, 93% H2SO4 solution (weight/volume).

It means 93 g of H2SO4 present in 100 mL of solution,

therefore, weight of H2SO4 = 93 g

Volume of solution = 100 mL

So, Weight of solution = 100 x 1.84 = 184 g

Thus, the weight of water =184 - 93=91 g

\therefore \text { Molality }=\frac{\text { moles of } \mathrm{H}_{2} \mathrm{SO}_{4}}{\text { weight of water in }(\mathrm{g})} \times 1000

\text { Moles of } \mathrm{H}_{2} \mathrm{SO}_{4}=\frac{93}{98}=0.948

\therefore \text { Molality }=\frac{0.948}{91} \times 1000=10.42

Posted by

shivangi.bhatnagar

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