Question

HW.9

How many grams of solute are in each of the following solutions?

Part A

4.5 L of a 1.4 M Al(NO3)3

Express your answer using two significant figures.

Part B

95 mL of a 0.90 M C6H12O6

Express your answer using two significant figures.

Part C

295 mL of a 1.80 M LiCl

Express your answer using three significant figures.

Answer 1

Part A

Concentration of Al(NO3)3 solution = 1.4 M = 1.4 mol/L

Volume of solution = 4.5 L

Molar mass of Al(NO3)3 = 213 g/mol

Mass of solute, Al(NO3)3 = 1.4 mol/L * 4.5 L * 213 g/mol = 1341.90 g

Part B

Concentration of C6H12O6 solution = 0.90 M = 0.90 mol/L

Volume of C6H12O6 solution = 95 ml = 95 L/1000 = 0.095 L

Molar mass of C6H12O6 = 180.16 g/mol

Mass of solute, C6H12O6 = 0.90 mol/L * 0.095 L * 180.16 g/mol = 15.40 g

Part C

Concentration of LiCl solution = 1.80 M = 1.80 mol/L

Volume of solution = 295 ml = 295 L/1000 = 0.295 L

Molar mass of LiCl = 42.4 g/mol

Mass of solute, LiCl = 1.80 mol/L * 0.295 L * 42.4 g/mol = 22.514 g