Calculate the molality of 12.5% w/w sulphuric acid?


Molality is defined as the “total moles of a solute contained in a kilogram of a solvent.


\(\begin{array}{l}Molality(M)= \frac{Number\, of\, moles\, of\, solute}{Mass\, of\, solvent\, in\, kgs}\end{array} \)


\(\begin{array}{l}Molality(M)= \frac{g \times 1000}{W \times m}\end{array} \)


12.5% w/w means 12.5 g in 100 g of solution.

Weight of solvent = 100 g – 12.5 g= 87.5 g.

Number of moles of sulphuric acid(H2SO4) = 12.5/ 98

Number of moles of H2SO= 0.127 mol


\(\begin{array}{l}Molality(M)= \frac{0.127\times 1000}{87.5}\end{array} \)


Molality (M) = 1.45 m

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