Question

How much dry solute would you take to prepare each of the following solutions from the dry solute and the solvent?

A. 119 mL of 0.120 M NaNO3

B. 122 g of 0.220 m NaNO3

C. 122 g of 1.4 % NaNO3 solution by mass

D. How much solvent would you take to prepare the solution in part B?

E. How much solvent would you take to prepare the solution in part C?

Answer 1

A. volume of solution = 119 mL

molarity of NaNO3 = 0.120M

moles of NaNO3 = 0.120*0.119

= 0.01428 moles

mass of NaNO3 = 0.01428moles * 84.9947 g/mol

= 0.123 grams

B. molality of solution = 0.220m

Mass of solution = 122 g

= 0.122 kg

moles of NaNO3 = 0.220*0.122

= 0.02684 moles

mass of NaNO3 =  0.02684 moles * 84.9947 g/mol

= 2.281 grams

C. mass % of NaNO3 = 1.4 %

mass of solution = 122 g

mass of NaNO3 = 1.4% of 122 g

= 0.014 * 122

= 1.708 g

D. mass of solution = 122 g

mass of solute = 2.281 g

mass of solvent = 122 - 2.281

= 119.719 g

E. mass of solution = 122 g

mass of solute = 1.708 g

mass of solvent  = 122 - 1.708

= 120.292 g