Question

the partial molar volumes of two A and B in a mixture in which the
MASS PERCENTAGE of A is 41.8% are 188.2 cm3/mol and 176.14 cm3/mol,
respectively. the molar masses of A and B are 241.1g/mol and 198.2
g/mol. what is the volume cm3 of a solution of mass 1.000 kg

Answer #1

**Answer** – Given, molar volume of A = 188.2
cm^{3}/mol

For B = 176.14 cm^{3}/mol

Mass percent of A = 41.8 % , mass of solution = 1.000 kg = 1000 g

Molar mass of A = 241.1g/mol and B = 198.2 g/mol

We need to calculate mass of A from the given mass percent

We know

Mas percent A = mass of A / total mass *100 %

So, mass of A = mass percent of A * total mass / 100 %

= 41.8 % * 1000 g / 100 %

= 418 g of A

So mass of B = 1000 g – 418 g

= 582 g

Moles of A = 418 g / 241.1 g.mol^{-1}

= 1.73 moles

Moles of B = 582 g / 198.2 g.mol^{-1}

= 2.94 moles

So for 1 moles of A = 188.2 cm^{3}

So, 1.73 mole = ?

= 326.3 cm^{3}

for 1 moles of B = 176.14 cm^{3}

So, 2.94 mole = ?

= 517.2 cm^{3}

So **total volume of solution** = 326.3
cm^{3} + 517.2 cm^{3}

=
**843.5 cm ^{3}**

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