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Practicing with Concentration Expressions. Molarity Percent ppm. White Board Review. Practicing with Concentrations. Molarity. Determine the molarity of 250 mL of solution containing 35 g of NaOH. What you must remember: molarity = moles of solute per liter of solution. - PowerPoint PPT Presentation
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White Board Review
Practicing with Concentration Expressions
•Molarity•Percent•ppm
Determine the molarity of 250 mL of solution containing 35 g of NaOH.
mol soluteM =
L solution
What you must remember: molarity = moles of solute per liter of solution
1 mol NaOH
40.0 g NaOH
=
35 g NaOH
0.250 Liters
= 3.5 M NaOH
Practicing with Concentrations Molarity
What volume of a 2.0 M solution will provide 18 g of NaOH?
1.0 L
2.0 mol NaOH
What you must remember: moles solute ÷ molarity = L volume
1 mol NaOH
40.0 g NaOH
=18 g NaOH 0.45 mol NaOH
= 0.23 L0.45 mol NaOH
or 230 mL of solution
First find moles:
Next, divide moles by molarity:
Practicing with Concentrations Molarity
What mass of NaOH is needed to make 4.0 liters of a 0.75 M solution?
40.0 g NaOH
1 mol NaOH
What you must remember: L volume X molarity = moles solute
0.75 mol NaOH
L solution
=4.0 L solution 3.0 mol NaOH
=120 g NaOH3.0 mol NaOH
First find moles solute:
Next, multiply moles by molar mass:
Practicing with Concentrations Molarity
Determine the molarity of 850 mL of solution containing 95 g of KCl.
mol soluteM =
L solution
What you must remember: molarity = moles of solute per liter of solution
1 mol KCl
74.6 g KCl
=
95 g KCl
0.850 Liters
= 1.5 M KCl
Practicing with Concentrations Molarity
What volume of a 3.8 M solution will provide 120 g of LiF?
1.0 L
3.8 mol LiF
What you must remember: moles solute ÷ molarity = L volume
1 mol LiF
25.9 g LiF
=120 g LiF 4.63 mol LiF
= 1.2 L4.63 mol LiF
First find moles:
Next, divide moles by molarity:
Practicing with Concentrations Molarity
Practicing with Concentrations Molarity
What mass of NaSCN is needed to make 550 mL of a 2.4 M solution?
81.1 g NaSCN
1 mol NaSCN
What you must remember: L volume X molarity = moles solute
2.4 mol NaSCN
L solution
=0.55L solution 1.32 mol NaSCN
=110 g NaSCN1.32 mol NaSCN
First find moles solute:
Next, multiply moles by molar mass:
Determine the % w/v of 25 mL of solution containing 3.0 g of CsOH.
mv
2g solute% = 10
mL solution
What you must remember: % w/v = g of solute per mL of solution X 100
= 3.0 g CsOH
25 mL
= 12 % w/v CsOH
Practicing with Concentrations Percent
x 102
What volume of a 6.2 % w/v solution will provide 32 g of KMnO4?
What you must remember: g solute ÷ % w/v = mL volume
Practicing with Concentrations Percent
4
solution
6.2 g KMnO
100 mL
= 520 mL KMnO4 solution
First interpret % w/v: Next, divide the g solute by the %:
32 g4
solution100 mL
6.2 g KMnO
Practicing with Concentrations
What mass of NaClO is needed to make 150 mL of a 4.9 % w/v solution?
What you must remember: % w/v = g/100mL and mL volume X % = g solute
solution
4.9 g NaClO
100 mL
= 7.4 g NaClOX
First interpret % w/v: Next, multiply by mL volume:
Percent
150 mL
Determine the % w/v of 960 mL of solution containing 37 g of KClO3.
mv
2g solute% = 10
mL solution
What you must remember: % w/v = g of solute per mL of solution X 100
= 37 g KClO3
960 mL
= 3.9 % w/v KClO3
Practicing with Concentrations Percent
x 102
What volume of a 0.55 % w/v solution will provide 2.5 g of KNO2?
What you must remember: g solute ÷ % w/v = mL volume
Practicing with Concentrations Percent
2
solution
0.55 g KNO
100 mL
= 450 mL KNO2 solution
First interpret % w/v: Next, divide the g solute by the %:
2.5 g2
solution100 mL
0.55 g KNO
Practicing with Concentrations
What mass of NaSO4 is needed to make 500. mL of a 1.0 % w/v solution?
What you must remember: % w/v = g/100mL and mL volume X % = g solute
4
solution
1.0 g NaSO
100 mL
= 5.0 g NaSO4X
First interpret % w/v: Next, multiply by mL volume:
Percent
500. mL
Calculate the concentration in ppm of 3250mL of solution containing 0.25 g of NiSeO3.
6g soluteppm = 10
mL solution
What you must remember: ppm = g of solute per mL of solution X 106
= 0.25 g NiSeO3
3250 mL
= 77 ppm NiSeO3
Practicing with Concentrations ppm
x 106
What volume of a 320 ppm solution will provide 0.75 g of K2Cr2O7?
What you must remember: ppm = mg/L and mg solute ÷ ppm = L volume
Practicing with Concentrations
2 2 7
solution
320 mg K Cr O
1 L
= 750 mg
First interpret ppm: Next, convert g to mg:
0.75 g 1000 mg
1 g
ppm
Finally, divide mg solute by ppm:
750 mg2 2 7
solution1 L
320 mg K Cr O
= 2.3 L
Practicing with Concentrations
What mass of MgSO3 is needed to make 100. mL of a 38 ppm solution?
What you must remember: ppm = mg/L and L volume X ppm = mg solute
3
solution
38 mg MgSO
1 L
= 3.8 mg MgSO3X
First interpret ppm: Next, multiply by L volume:
0.100 L
ppm
Calculate the concentration in ppm of 25,000mL of solution containing 1.0 g of KF.
6g soluteppm = 10
mL solution
What you must remember: ppm = g of solute per mL of solution X 106
= 1.0 g KF
25,000 mL
= 40. ppm KF
Practicing with Concentrations ppm
x 106
What volume of a 5.0 ppm solution will provide 1.0 g of NaCN?
What you must remember: ppm = mg/L and mg solute ÷ ppm = L volume
Practicing with Concentrations
solution
5.0 mg NaCN
1 L
= 1000 mg
First interpret ppm: Next, convert g to mg:
1.0 g 1000 mg
1 g
ppm
Finally, divide mg solute by ppm:
1000 mgsolution1 L
5.0 mg NaCN
= 2.0 x 102 L
Practicing with Concentrations
What mass of FePO4 is needed to make 500. mL of a 120 ppm solution? What you must remember:
ppm = mg/L and L volume X ppm = mg solute
4
solution
120 mg FePO
1 L
= 60. mg FePO4X
First interpret ppm: Next, multiply by L volume:
0.500 L
ppm
