Question

Molarity of h2so4 is 0.8 m and its density is 1.06 g cmââ¬â3. what will be the concentration of the solution in terms of molality (m) and mole fraction (x)

Answer 1

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Answer 2

Answer:

Molality of the solution = 0.815 moles and mole fraction = 0.985

Explanation:

We are given that the molarity of $\bold{\mathrm{H}_{2} \mathrm{SO}_{4}}$ is 0.8. This means that 0.8 moles of H2SO4 are present in 1000 ml of the solution. So, the number of moles of solute=0.8.

The mass of the solution=1000ml

Density of the solution=1.06 gm (given)

Also Density= Mass of solution ÷ Volume of the solution

   1.06= Mass of solution ÷ 1000

So, we get the mass of solution= 1060 gm

The molar weight of $\mathrm{H}_{2} \mathrm{SO}_{4}$= 98 g/mol

Mass of 0.8 mol of $\mathrm{H}_{2} \mathrm{SO}_{4}$= 0.8mol×98g/mol =78.4gm

So, now we have the total mass of the solution= 1060 gm and the mass of solute= 78.4 gm

So, the mass of solvent= (1060-78.4)gm= 981.6 gm

So, the mass of water= 981.6gm

The molar weight of water= 18g/mol

We can now calculate the moles of water using the formula:

Number of moles of water= Mass of water ÷ Molar weight of water

=981.6gm ÷ 18g/mol= 54.53 mol

The molality of the solution= (Mass of solute÷ Mass of solvent) ×1000

=(0.8÷981.6×1000=0.815 moles

The mole fraction of $\mathrm{H}_{2} \mathrm{SO}_{4}$= 0.8/(0.8+54.53)= 0.0145

The moles fraction of water= 54.53(0.8+54.53)=0.985