Question

A solution is made by dissolving 27.8 g of nicotine (molar mass= 160 g/mol) in 145 g of cyclohexane (C₆H₁₂) to form 142 mL of solution. Calculate the m ole fraction of the solute and the molarity, respectively, of this solution.

a. 0.174, 1.22 M

b. 0.101, 12.1 M

c. 0.101, 1.22 M

d. 0.091, 1.22 M

e. 0.174, 12.1 M

Answer 1

The answer is   d

Explanation:

Mole fraction of the solute = n₂ /n₁+n₂ where n₁ is the number of moles of the solvent and n₂, the number of moles of the solute

n₁ = mass of the solvent (cyclohexane C₆H_12, 145 g) / Molar mass of solvent (6X12 + 12x1 =84)

    = 145 g/84 g = 1.7261

n_{2 =} mass of the solute (nicotine, 27.8g)/ Molar mass of solute (160g)

    = 27.8g/160g = 0.1737

Total number of moles n₁+n₂ = 1.7261+0.1737 = 1.8998

Molefraction of the solute, nicotine= 0.1737/1.8998= 0.0914

Molarity is the number of moles of the solute present in 1000 ml of the solution

142 ml of solution contains 0.1735 moles of solute

Therefore 1000 ml of the solution will contain 0.1735x1000/142 = 1.22 moles of the solute

Molarity of the solution = 1.22 M

Answer: d 0.091, 1.22 M