Respuesta :
Explanation:
The given data is as follows.
Weight of solute = 75.8 g, Molecular weight of solute (toulene) = 92.13 g/mol, volume = 200 ml
- Therefore, molarity of toulene is calculated as follows.
Molarity = [tex]\frac{\text{weight of solute}}{\text{molecular weight of solute}} \times \frac{1000}{\text{volume of solution in ml}}[/tex]
= [tex]\frac{75.8 g}{92.13 g/mol} \times \frac{1000}{200 ml}[/tex]
= 4.11 M
Hence, molarity of toulene is 4.11 M.
- As molality is the number of moles of solute present in kg of solvent.
So, we will calculate the molality of toulene as follows.
Molality = [tex]\frac{\text{given weight of solute}}{\text{given molecular weight of solute}} \times \frac{1000}{\text{weight of solvent in grams}}[/tex]
= [tex]\frac{75.8 g}{92.13 g/mol} \times \frac{1000}{95.6 g}[/tex]
= 8.6 m
Hence, molality of given toulene solution is 8.6 m.
- Now, calculate the number of moles of toulene as follows.
No. of moles = [tex]\frac{mass}{\text{molar mass}}[/tex]
= [tex]\frac{75.8 g}{92.13 g/mol}[/tex]
= 0.8227 mol
Now, no. of moles of benzene will be as follows.
No. of moles = [tex]\frac{mass}{\text{molar mass}}[/tex]
= [tex]\frac{95.6 g}{78.11 g/mol}[/tex]
= 1.2239 mol
Hence, the mole fraction of toulene is as follows.
Mole fraction = [tex]\frac{\text{moles of toulene}}{\text{total moles}}[/tex]
= [tex]\frac{0.8227 mol}{(0.8227 + 1.2239) mol}[/tex]
= 0.402
Hence, mole fraction of toulene is 0.402.
- As density of given solution is 0.857 [tex]g/cm^{3}[/tex] so, we will calculate the mass of solution as follows.
Density = [tex]\frac{mass}{volume}[/tex]
0.857 [tex]g/cm^{3}[/tex] = [tex]\frac{mass}{200 ml}[/tex] (As 1 [tex]cm^{3}[/tex] = 1 g)
mass = 171.4 g
Therefore, calculate the mass percent of toulene as follows.
Mass % = [tex]\frac{\text{mass of solute}}{\text{mass of solution}} \times 100[/tex]
= [tex]\frac{75.8 g}{171.4 g} \times 100[/tex]
= 44.22%
Therefore, mass percent of toulene is 44.22%.
