Calculate the molarity of the solutions described below.

A. 10.0 g of sodium chloride is dissolved in 2.0 L of solution.

B. 52.5 g of sugar (C12H22O11) s dissolved in 1.0 L of solution.

C. 120 g of aluminum sulfate is dissolved in 10.0 L of solution.

D. 1.75 g of caffeine (C8H10N4O2) is dissolved in 0.100 L of solution.

Question

Chemistry

Isabelle Benson

9 March, 17:10

Calculate the molarity of the solutions described below.

A. 10.0 g of sodium chloride is dissolved in 2.0 L of solution.

B. 52.5 g of sugar (C12H22O11) s dissolved in 1.0 L of solution.

C. 120 g of aluminum sulfate is dissolved in 10.0 L of solution.

D. 1.75 g of caffeine (C8H10N4O2) is dissolved in 0.100 L of solution.

+2

Answers (1)

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Devin Guerra · Answer 1

A. Na = 23 g/m Cl = 35.5 g/m NaCl = 58.5 g/m

10g / 58.5 g/m =.17 mole

.17 mole / 2.0 L = 0.086 m/L = 0.086 Molar

B. 342 g/m 52.5/342 = 0.15 mole 0.15 mole/1.0 L = 0.15 Molar

C. Al2 (SO4) 3= = 342 g/m 120/342 = 0.35 mole 0.35 m/10. L 0.035 Molar

D caffeine = 194 g/m 1.75 / 194 = 0.009 mole 0.009/0.100 L 0.09 Molar