Answer to Question #159601 in General Chemistry for Lizzy

Question

Read, analyze, and answer the problems below. Determine the Molarity, Molality, and Mole Fraction

1.  A solution is used for intravenous feeding contains 5.60g of glucose in 100g of water.

Data

Molar Mass

*Glucose = 180.16 g/mol

*Water = 18 g/mol

Density of Solution = 3.52 g/mL

a. Molality of glucose

b. Molarity of glucose

c. Mole fractions of glucose and water

d. Percent by mass of glucose

Answer 1

a) Molality = 5.6/100 x 0.18 = 0,311111111.

b) Molarity = 5.6 x 1000 x 3.52/100 x 180 = 1,09511111 M.

c) 5.6/180 = 0,0311111111 mol.

100/18 = 5,55555556 moles.

Mol fraction = 0.0311111 x 100/5.55556 = 0,5568814 % of glucose.

100-0,5568814 = 99,4431186 %. Of water.

d) 5.6 x 100%/ (5.6 + 100) = 5,3030303%.

100-5.3030303 = 94,6969697%.