Answers for Homework on Molarity due for Friday, January 11th, 2013.
p. 482
#10.
How many moles of ammonium nitrate are in 335 mL of 0.425 M NH 4NO3?
0.425 mol NH4NO3 x __1 L __ x 335 mL
L solution
1000 mL
#12.
=
0.140 mol NH4NO3
How many milliliters of a solution of 4.00 M KI are needed to prepare
250.0 mL of 0.760 MKI?
Diluting the 4.00 M KI to
M1
V1
0.760 M KI determining the amount of water needed.
= M2 V2
(0.760 M) (250.0 mL) =
(4.00 M)
V2
solve for the unknown volume V2
47.50 mL
= 47.50 mL of the 4.00 M of KI should be placed into the 250 mL
volumetric flask. After that is in the flask water is added
until it reaches the etched line (total volume of 250.0 mL).
#19.
Calculate the molarity of a solution containing 400 g CuSO4 in 4.00 L of solution.
400 g CuSO4 x 1 mol CuSO4 =
4.00 L
159.6 g CuSO4
#20.
0.630 mol CuSO4
L (solution total)
=
0.630 M CuSO4
How many moles of solute are present in 50.0 mL of 0.20 M KNO 3?
0.20 M KNO3 remember that M = mol solute (KNO3)
L solution
0.20 mol solute (KNO3)
L solution
x
1L x
1000mL
50.0 mL
=
0.0100 mol KNO3