Question

Calculate the volume in mL of a solution required to provide the following.

(a) 2.14 g of sodium chloride from a 0.180 M solution mL

(b) 4.10 g of ethanol from a 1.50 M solution mL

(c) 0.85 g of acetic acid (CH3COOH) from a 0.50 M solution mL

Answer 1

a)

number of moles of NaCl = 2.14g / 58.44 g/mol = 0.0366 mole

molarity = number of moles / volume of solution in L

0.180 = 0.0366 mole / volume of solution in L

volume of solution inL = 0.0366 / 0.180 = 0.203 L

1 L = 1000 mL

0.203 L = 203 mL

Therefore, the volume of solution required is 203 mL

b)

number of moles of NaCl = 4.10g / 46.068 g/mol = 0.0890 mole

molarity = number of moles / volume of solution in L

1.50 = 0.0890 mole / volume of solution in L

volume of solution in L = 0.0890 / 1.50 = 0.0593L

1 L = 1000 mL

0.0593 L = 59.3 mL

Therefore, the volume of solution required is 59.3 mL

c)

number of moles of acetic acid = 0.85 g / 60.05 g/mol = 0.0142 mole

molarity = number of moles / volume of solution in L

0.50 = 0.0142 mole / volume of solution in L

volume of solution inL = 0.0142 / 0.50 = 0.0284 L

1 L = 1000 mL

0.0284 L = 28.4 mL

Therefore, the volume of solution required is 28.4 mL