Answer :

strawberry4300
(0.058 mol I2)(253.808 g/mol I2) = 14.721 g of I2 
So that's the mass of I2: 14.721 g 
Percentage is the mass of solute/(mass of solute + mass of solvent) 

14.721/(14.721g + 105 g) 
14.721/119.721 
=0.123 or 12.3%

Answer: The mass percent of iodine in the solution is 6.77 %

Explanation:

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

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]

We are given:

Moles of iodine gas = [tex]4.0\times 10^{-2}=0.04moles[/tex]

Molar mass of iodine gas = 254 g/mol

Putting values in above equation, we get:

[tex]0.04mol=\frac{\text{Mass of iodine gas}}{254g/mol}\\\\\text{Mass of iodine gas}=(0.04mol\times 254g/mol)=10.16g[/tex]

To calculate the percentage composition of a substance in a solution, we use the equation:

[tex]\%\text{ composition of substance}=\frac{\text{Mass of substance}}{\text{Mass of solution}}\times 100[/tex]

We are given:

Mass of solute (iodine gas) = 10.16 g

Mass of solvent (carbon tetrachloride) = 140 g

Mass of solution = (140 + 10.16) g = 150.16 g

Putting values in above equation, we get:

[tex]\text{Mass percent of iodine in solution}=\frac{10.16}{150.16}\times 100\\\\\text{Mass percent of iodine in solution}=6.77\%[/tex]

Hence, the mass percent of iodine in the solution is 6.77 %

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