`4.50g` sulphuric acid was added to `82.20g` water and the density of the solution was found to be `1.029g//`cc at `25^(@)C` and 1 atm pressure. Calcu

Question

`4.50g` sulphuric acid was added to `82.20g` water and the density of the solution was found to be `1.029g//`cc at `25^(@)C` and 1 atm pressure. Calcu

`4.50g` sulphuric acid was added to `82.20g` water and the density of the solution was found to be `1.029g//`cc at `25^(@)C` and 1 atm pressure. Calculate (a) the weight percent, (b) the mole fraction, (c) the mole percent, (d) the molality, (e) the molarity of sulphric acid in the solution under these conditions .

Monjur Alahi

8 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Sulphuric acid = 4.450 g, Water = `82.20g rArr Wt`. of solution = 86.65 g
`:.` Density of solution = 1.029 g/cc
(a) Weigth percent `= ("wt. of solute")/("wt. of solution") xx 100 = (4.450)/(86.65) xx 100 = 5.14`
(b) Mole fraction:
Mole of solute `= ("wt. of solute")/("mol wt. of solute") = (4.45)/(98) = 0.0454`
Mole of solvent `= (82.20)/(18) = 4.566`
Total moles in solution `= 0.0454 + 4.566 = 4.6114`
Mole fraction of solute `= (0.0454)/(4.6114) = 0.0098`
(c0 Mole percent `= ("moles of solute")/("Total moles in solution") xx 100`
= mole fraction of solute `xx 100 = 0.0098 xx 100 = 0.98`
(d) Molality `= ("moles of solute")/("mass of solvent(in gm)") xx 1000`
`= (0.0454 xx 1000)/(82.2) = 0.552`
(e) Molarity `= ("moles of solute")/("litre of solution")`
Volume of solution `= ("Mass")/("Density") = (86.65)/(1.029) ml`
`= (86.65)/(1.029 xx 1000)` litre
Molarity `= (0.0454)/(86.54) = (0.0454 xx 1000 xx 1.029)/(86.65) = 0.539`
`1.029 xx 1000`