`4.50g 100` per 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 .

Explanation

Sulphuric acid `=4.450g` Water `=82.20 g implies "Wt of solution" =86.65g`
`:. "Density of solution" =1.029 g//c c`
(a) Weight percent ` = ("wt. of solute")/("wt .of solution") xx 100 =(4.450)/(86.65)xx100=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)/(13) = 4.566`
Total moles in solution `= 0.0454 + 4.566 = 46114`
Mole fractions of solute `= (0.0454)/(4.6114) = 0.0098`
(c) Mole percent `= ("moles of solute")/("Total moles in solution") xx 100`
= mole fractions of solute `xx 100 = 0.0098 xx 100 = 0.98`
(d) Molality = `("moles of solute")/("mass of solvent (ingm)") 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/(1.029xx1000))-(0.0454 xx 1000 xx 1.029)/(86.65)=0.539` .