Question

Part A:

How much dry solute would you take to prepare 127 mL of 0.120 M NaNO3?

Part B:

How much dry solute would you take to prepare130 g of 0.100 m NaNO3?

Part C:

How much dry solute would you take to prepare 130 g of 2.0 % NaNO3 solution by mass?

Part D:

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

Part E:

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

Answer 1

A)

volume , V = 1.27*10^2 mL

= 0.127 L

use:

number of mol,

n = Molarity * Volume

= 0.12*0.127

= 1.524*10^-2 mol

Molar mass of NaNO3,

MM = 1*MM(Na) + 1*MM(N) + 3*MM(O)

= 1*22.99 + 1*14.01 + 3*16.0

= 85 g/mol

use:

mass of NaNO3,

m = number of mol * molar mass

= 1.524*10^-2 mol * 85 g/mol

= 1.295 g

Answer: 1.30 g

B)

m(solvent)= 130 g

= 0.13 kg

use:

number of mol,

n = Molality * mass of solvent in Kg

= (0.1 mol/Kg)*(0.13 Kg)

= 1.3*10^-2 mol

Molar mass of NaNO3,

MM = 1*MM(Na) + 1*MM(N) + 3*MM(O)

= 1*22.99 + 1*14.01 + 3*16.0

= 85 g/mol

use:

mass of NaNO3,

m = number of mol * molar mass

= 1.3*10^-2 mol * 85 g/mol

= 1.105 g

Answer: 1.11 g

c)

mass % = mass of solute * 100 / mass of solution

2.0 = mass of solute * 100 / 130

mass of solute = 2.60 g

Answer: 2.60 g

D)

mass of NaNO3 = 1.11 g

mass of solution = 130 g

so,

mass of solvent = mass of solution - mass of NaNO3

= 130 g - 1.11 g

= 129 g

Answer: 129 g

E)

mass of NaNO3 = 2.60 g

mass of solution = 130 g

so,

mass of solvent = mass of solution - mass of NaNO3

= 130 g - 2.60 g

= 127 g

Answer: 127 g