Question

You want to make 750.00 mL of a 0.9550 M aqueous solution of Cr(NO3)2. You have available in lab a bottle of pure Cr(NO3)2 ∙ 6 H2O (s), a bottle of 1.672 M Cr(NO3)2 solution and a bottle of 0.5005 M Cr(NO3)2.

a) How many grams of the Cr(NO3)2 ∙ 6 H2O (s) would you need to use to prepare the desired solution?

b) How many mL of the 1.672 M Cr(NO3)2 solution would you need to use to prepare the desired solution?

c) Describe how you could use the 0.5005 M Cr(NO3)2 solution to prepare the desired solution.

Answer 1

a) How many grams of the Cr(NO3)2 ∙ 6 H2O (s) would you need to use to prepare the desired solution?

First calculate the moles of Cr(NO3)2 in 0.9550 M

Number of moles = molarity * volume in L

= 0.9550 M*0.750 L

= 0.71625 moles

1 mole of Cr(NO3)2 ∙ 6 H2O (s) have one mole of Cr(NO3)2

Thus,

0.71625 moles Cr(NO3)2 ∙ 6 H2O (s) have 0.71625 moles of Cr(NO3)2

Amount of Cr(NO3)2 ∙ 6 H2O (s) = number of moles * molar mass

= 0.71625 moles Cr(NO3)2 ∙ 6 H2O (s) *346.1025 g/mol

= 247.9 g

b) How many mL of the 1.672 M Cr(NO3)2 solution would you need to use to prepare the desired solution?

M1V1= M2V2

Here M1= 1.672 M

V2 = 750.00 mL M2 = 0.9550 M

Then;

M1V1= M2V2

1.672 *V2= 0.9550*750 ml

V2 = 428.38 ml

c) Describe how you could use the 0.5005 M Cr(NO3)2 solution to prepare the desired solution.

M1V1= M2V2

Here M1= 0.5005 M

V2 = 750.00 mL M2 = 0.9550 M

Then;

M1V1= M2V2

0.5005 *V2= 0.9550*750 ml

V2 = 1431.01 ml