Question

Mole fraction of i2 in c6 h6 is 0.2 calculate the molality of i2 in c6 h6

Answer 1

Let us start of with an assumption: we have ten moles of the given solution. Therefore we have 2 moles of iodine dissolved in 8 moles of benzene.

Molecular mass of benzene=78.

Molality=No.molesofiodine/MassofsolventinkgNo.molesofiodine/Massofsolventinkg

Therefore 8 moles of benzene weighs 78∗8=624grams78∗8=624grams

Molality=(2∗1000)/624=(2∗1000)/624

= 3.205

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Answer 2

The molality of iodine in benzene solution is 3.205 m

Explanation:

We are given:

Mole fraction of iodine = 0.2

This means that 0.2 moles of iodine is present in 1 mole of a solution

Moles of benzene = (1 - 0.2) = 0.8 moles

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of benzene = 0.8 moles

Molar mass of benzene = 78 g/mol

Putting values in above equation, we get:

$0.8mol=\frac{\text{Mass of benzene}}{78g/mol}\\\\\text{Mass of benzene}=(0.8mol\times 78g/mol)=62.4g$

To calculate the molality of solution, we use the equation:

$\text{Molality}=\frac{n_{solute}\times 1000}{W_{solvent}\text{ (in grams)}}$

where,

$n_{solute}$ = Moles of solute (iodine gas) = 0.2 moles

$W_{solvent}$ = Mass of solvent (benzene) = 62.4 g

Putting values in above equation, we get:

$\text{Molality of solution}=\frac{0.2\times 1000}{62.4}\\\\\text{Molality of solution}=3.205m$

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