Directions: Solve the following molarity problems, i.e. describe how you would p
ID: 482810 • Letter: D
Question
Directions: Solve the following molarity problems, i.e. describe how you would prepare the following solutions, dive the mass and the moles of the solute needed. 600 mL of2.0 M NaCI 275 ml. of0.250 M KNO, 350 mL of 1.5 M CuSO_4 5.00 L of 0.100 M Ca(OH)_2 325 mL of 1.25 M (NH_4)_2SO_4 650 mL of 0.0956 M LiNO_3 Direction: Calculate the molarity of the following solutions from the given information 405 g of NaOH in 2.0 L 155 g of Fe_2O_3 in 550 ml. 550 g of (NH_4)_2SO_4 in 750 ml 89.5 g of LiOH in 250 ml. 92.5 g of Al_2(SO_4)_3 in 500 mL 95.8 go of NaBr in 625 mL Directions: Solve the following dilution problems. Determine the concentration of the resulting solution when 267 ml of 3.2 M NaBr is mixed with 233 mL of water. How much water must he added to 550 ml. of 2.0 M HCI to prepare a solution of 0.75 M HCI? How much water must be added to 350 mL of 3.5 M Ca(OH)_2 to prepare a solution of 2.0 M Ca(OH)_2?Explanation / Answer
Answer:
1) 600 mL of 2M NaCl:
NaCl molecular weight = 58.44
Molarity = (weight of solute / molecular weight ) x (1000/ V in mL)
2 = (wt / 58.44) x (1000/600)
2 = (wt / 58.44) x (1.666)
1.666 wt = 2 x 58.44
wt = 116.88/1.666
wt = 70.156 g
Hence to prepare the 2M of 600 mL NaCl, We have to dissolve 70.1 g of NaCl in 600 mL of spolvent.
7) 405 g of NaOH in 2L :
NaOH molecular weight = 40
Molarity = (weight of solute / molecular weight ) x (1000/ V in mL)
M = (405 / 40) x (1000/2000)
M = 5.06
Hence the Molarity = 5.06 M.
13) 267 mL of 3.2 M NaBr is mixed with 233 mL of water:
M1 = 3.2 V1 = 267 mL
Water added (V2) = 233 mL.
Resultant concentration:
MV = M1V1 V =Total volume
V = V1 + V2
V = 267 + 233
V = 500 mL
MV = M1V1
M x 500 = 3.2 x 267
M = 854.4 / 500
M = 1.708
Hence the concentration after the dilution = 1.708M
14) How much water added 550 mL 2M HCl to prepare a solution of 0.75 M HCl:
V1 = 550 mL , M1 = 2M
V2 = ? M2 = 0.75 M
M2V2 = M1V1
0.75 x V2 = 2 x 550
V2 = 1100 / 0.75
V2 = 1466.66 mL
The addition of the water = Total volume(V2) - Initial volume(V1)
Water volume = 1466.66 - 550
= 916.66 mL.
Hence we have to add 916.66 mL of water.
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