Question

26. What would be the Molarity of an aqueous solution that has a mole fraction of glucose, C6H12O6 equal to 0.183? The density of the solution is 1.32 g/mL. a. 5.07 M b. 3.52 M c. 1.86 M do 0.723 M

Answer 1

Answer

a) 5.07 M

Explanation

Consider 1L of solution

density of solution = 1.32g/ml

mass of solution = 1000ml × 1.32g/ml = 1320g

mole fraction of H2O + mole fraction of glucose = 1

x + 0.183 = 1

x = 0.817

mass fraction = mole fraction × ( Molar mass of the component) / (weight avarage molar mass of the mixture)

weight average molar mass of mixture = ( 0.183 × 180.16g/mol )+ (0.817×18.016g/mol) = 32.969g/mol + 14.719g/mol = 47.688g/mol

mass fraction of glucose = 0.183 × (180.16g/mol/47.688g/mol) = 0.6914

mass of glucose = 0.6914 × 1320g = 912.64g

number of moles of glucose = 912.64g/180.16g/mol = 5.07 mol

molarity is defined as number of moles of solute per liter of solution

Therefore

Molarity of the solution = 5.07M