A solution is created by adding 10.0 g of octane (C_8H_8 MM = 114g/mole) in 90.0

Question

A solution is created by adding 10.0 g of octane (C_8H_8 MM = 114g/mole) in 90.0g of benzene (C_6H_6 MM = 78.0g/mole). The mixture has a density of 0.720g/ml, calculate the: Molality, Mass percentage of solute, Molarity. These two liquid mix very well together what type of intermolecular force is involved in allowing them to mix so well? Benzene has a boiling point of 116 degree F and octane doesn't boil until 257 degree F. Use the relative strengths of their IMF's to explain this difference in boiling points. Would you expect either of these liquids to dissolve in water? Why or why not?

Explanation / Answer

Given

Mass of solute ( octane) = 10 g

molar Mass of solute ( octane) = 114 g/mol

Mass of solvent ( benzene) = 90 g

Molar Mass of solvent ( benzene) = 78 g/mol

Density of Mixture (solution) = 0.720 g/ml

Solution:

(a) Molality = No. of moles of solute /Mass of solvent in Kg

No. of moles of solute = Mass of solute / molar mass of solute = 10 g /114 g/mol =0.0877 moles

Molality = 0.0877 moles / 90 g = 0.0877 moles / 0.09 Kg = 0.97 mol/kg

Molality = 0.97 m Answer

(b) mass % of solute = mass of solute * 100 %/ Mass of solution (solute +solvent) = 10 g * 100 %/ (90+10) g

mass % of solute = 10 % Answer

(c) Molarity = No. of moles of solute / Volume of solution in L

Volume of solution = mass of solution / density = (10 + 90) g / 0.72 g/ml = 138 mL = 0.138 L

molarity = 0.0877 moles / 0.138 L = 0.63 mol/L

Molarity = 0.63 M Answer

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.