Question

A Sulphuric acid solution contains 98% (m/v) H2SO4. The density of the solution is 1.98gcm^-3. The molarity and molality of the solution are

Answer 1

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=) Given ,
Weight of H2SO4 = 98 g .
Density = 1.98 g/cm3 .

we know , Weight of solution = 100 gm .
Solvent = water .

Acc . to question ...

.•. Volume of Solution = 100 ÷ 1.98 = 50.5 cm3.

Weight of water = 100 - 98 = 2 g .

Molecular weight of H2SO4 = 98 g/mol .

No• of moles in 98 gm = Given mass / molar mass .
= m/M = 98/98 = 1 mol .

Formula of Molality ,

=
$\frac{number \: of \: moles \: of \: solute \: }{weight \: of \: solvent} \times 100$
$\frac{1}{2} \times 100$
500 mol/kg .

Formula of Molarity ,

=
$\frac{number \: of \: moles \: of \: solute \: }{weightofsolution } \times 1000$
$\frac{1}{50.5} \times 1000$
19.8 mol .


Hence ,

Molality =500 mol/kg .

Molarity = 19.8 mol/ l .

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Answer 2

The molarity is 0.1M and molality is 10m