How many grams of NaCl (MW=58.4 g/mol) are contained in 350.0 mL of a 0.250 M so
ID: 785006 • Letter: H
Question
How many grams of NaCl (MW=58.4 g/mol) are contained in 350.0 mL of a 0.250 M solution of sodium chloride?
5.11 g
14.6 g
41.7 g
81.8 g
87.5 g
If 5.15 g FeCl3 is dissolved in enough water to make exactly 150.0 mL of solution, what is the molar concentration of chloride ion? The formula weight of FeCl3 is 162.4 g/mol) and the solution has a density of 1.2 g/mL.
0.103 M
0.212 M
0.578 M
0.635 M
16.7 M
If 5.00 mL of 0.314 M KOH is diluted to exactly 125 mL with water, what is the concentration of the resulting solution?
0.127 M
0.281 M
7.85 M
1.26 x 10-2 M
5.02 x 10-4 M
How many grams of NaCl (MW=58.4 g/mol) are contained in 350.0 mL of a 0.250 M solution of sodium chloride? If 5.15 g FeCl3 is dissolved in enough water to make exactly 150.0 mL of solution, what is the molar concentration of chloride ion? The formula weight of FeCl3 is 162.4 g/mol) and the solution has a density of 1.2 g/mL. If 5.00 mL of 0.314 M KOH is diluted to exactly 125 mL with water, what is the concentration of the resulting solution?Explanation / Answer
1)Moles = Molarity x Volume ( in Liters)
Moles = 0.250 x 0.350 L = 0.0875
Mass = moles x molar mass = 0.0875 mol x 58.4428 g/mol =5.11 g
2)Molar concentration = moles of solute/liters of solution
First change 5.15g FeCl3 to moles FeCl3 = 0.0318mol FeCl3
Figure out the balanced eqn for dissolved FeCl3
FeCl3 ? Fe+ + 3Cl-
Using the balanced eqn, find moles of Cl-
(0.0318mol FeCl3)(3mol Cl- / 1mol FeCl3) = 0.0954mol Cl-
Now use moles of Cl- divided by the volume in liters of solution to find molar concentration
(0.0954mol/.150L) = 0.635M
3)MV=MV
(0.314)(5) =M(125)
M=0.01256 M = 1.26 * 10^-2 M
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