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Calculations for Solutions Worksheet and Key, Lecture notes of Chemistry

Facultés Universitaires Notre-Dame de la PaixChemistry

Calculations for Solutions Worksheet and Key. 1) 23.5g of NaCl is dissolved in enough water to make .683 L of solution. a) What is the molarity (M) of the ...

Typology: Lecture notes

2021/2022

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Calculations+for+Solutions+Worksheet+and+Key+
1)++23.5g+of+NaCl+is+dissolved+in+enough+water+to+make+.683+L+of+solution.+
a)+What+is+the+molarity)(M)+of+the+solution?+
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b)++How+many+moles+of+NaCl+are+contained+in+0.0100+L+of+the+above+NaCl+solution?+
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c)++What+volume)(L)+of+this+NaCl+solution+would+contain+0.200+moles+of+NaCl?+
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2)++12.5g+of+glucose+(C6H12O6)++is+dissolved+in+enough+water+to+make+750.0+mL+of+solution.+
a)+What+is+the+molarity)(M))of+the+solution?+
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b)++How+many+moles+of+glucose+are+contained+in+237+mL+of+the+above+glucose+solution?+
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c)++What+volume)(L)+of+this+glucose+solution+would+contain+0.079+moles+of+glucose?+
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Calculations for Solutions Worksheet and Key

  1. 23.5g of NaCl is dissolved in enough water to make .683 L of solution. a) What is the molarity (M) of the solution? b) How many moles of NaCl are contained in 0.010 0 L of the above NaCl solution? c) What volume (L) of this NaCl solution would contain 0.200 moles of NaCl?
  2. 12.5g of glucose (C 6 H 12 O 6 ) is dissolved in enough water to make 750.0 mL of solution. a) What is the molarity (M) of the solution? b) How many moles of glucose are contained in 237 mL of the above glucose solution? c) What volume (L) of this glucose solution would contain 0.079 moles of glucose?
  1. 45.7 g of magnesium chloride (MgCl 2 ) is dissolved in 2.40 kg of water. a) What is the molality ( m ) of the solution? b) How many moles of MgCl 2 are contained in 1.76 kg of solvent? c) How many kg of solvent would contain 0.0150 moles of MgCl 2? 4 ) 114.5 g of KCl is dissolved in enough water to make 3.6 L of solution. a) How many osmoles are in one mole of KCl when it dissolves? b) What is the osmolarity of the solution? c) How many osmoles are contained in 1.00 L of the above potassium chloride solution? d) How many liters (L) of this potassium chloride solution would contain 0.3 50 osmoles?
  1. 234.5 g of KCl is dissolved in enough water to make 3.6 L of solution. a) How many equivalents of potassium (K+) are in one mole of KCl when it dissolves? (note: you are concerned with the Eq from K+^ only , do not include Eq from Cl-­‐ ) b) What is the concentration of potassium in (Eq/L)? c) How many equivalents (Eq) of K+^ are contained in 0.70 0 L of the above potassium chloride solution? d) How many liters (L) of this potassium chloride solution would contain 0. 050 equivalents Eq of K+?
  2. 0.250 g of aluminum sulfate is dissolved in enough water to make 150 mL of solution. a) How many equivalents of sulfate ion (SO 42 -­‐) are in one mole of aluminum sulfate when it dissolves? (note: you are concerned with the Eq from SO 42 -­‐^ only , do not include Eq from Al+ ) b) What is the concentration of sulfate in (Eq/L)? c) How many equivalents (Eq) of SO 42 -­‐^ are contained in 0.0280 L of the above aluminum sulfate solution? d) How many liters (L) of this aluminum sulfate solution would contain 0.0025 equivalents Eq of SO 42 -­‐?

Molarity calculations (fill-­‐in all the boxes)

solute moles of

solute

grams of

solute

volume of

solution

Concentration

(Molarity, M=mole/L)

NaCl 3.00 moles 0 .500 L

NaCl 13.5 g 0.150 L

NaCl 0 .375 moles 1.00 M

NaCl 0 .059 g 0.30 M

KNO 3 1.57 moles 0 .770 M

KNO 3 1.98 g 2.00 M

KNO 3 0 .288 L 0 .197 M

Osmolarity calculations

solute moles of

solute

osmoles of

solute

grams of

solute

volume of

solution

Concentration

(Osmolar = osmole/L )

KCl 2. 4 0 moles 0. 6 00 L

KCl 1 .5 g 0. 750 L

KCl 0. 0 50 moles 1.00 osmolar

KCl 0. 892 g 0. 150 osmolar

glucose

C 6 H 12 O 6

1.50 moles 1.22 osmolar

glucose

C 6 H 12 O 6

1.17 g 0 .0 100 osmolar

glucose

C 6 H 12 O 6

0. 375 L 0.0750 osmolar

c) What volume (L) of this glucose solution would contain 0.079 moles of glucose? 0.079 moles glucose L of solution = 0.85 L of solution 0.0925 moles glucose

  1. 45.7 g of magnesium chloride (MgCl 2 ) is dissolved in 2.40 kg of water. a) What is the molality ( m ) of the solution? Molar mass of MgCl 2 = 95.21 g/mole Moles of MgCl 2 : 45.7 g MgCl 2 1 mole MgCl 2 = 0.480 moles MgCl 2 95.21 g MgCl 2 Molality = moles = 0.480 moles MgCl 2 = 0.200 moles MgCl 2 /kg = 0. 200 m MgCl 2 kg of solvent 2.40 kg of solvent b) How many moles of MgCl 2 are contained in 1.76 kg of solvent? 1.76 kg solvent 0.200 moles MgCl 2 = 0.352 moles MgCl 2 1 kg of solvent c) How many kg of solvent would contain 0.0150 moles of MgCl 2? 0.0150 moles MgCl 2 1 kg of solvent = 0.0750 kg of solvent 0.200 moles MgCl 2 4 ) 114.5 g of KCl is dissolved in enough water to make 3.6 L of solution. a) How many osmoles are in one mole of KCl when it dissolves? one mole of KCl = 2 osmoles
  • This relationship can be used as a conversion factor to convert between moles and osmoles : 2 osmoles or 1 mole KCl 1 mole KCl 2 osmoles Concentration of the solution Concentration of the solution

b) What is the osmolarity of the solution?

  • First get the moles of KCl then convert to osmoles: Molar mass of KCl = 74.55 g/mole
  • Osmoles in solution: 114.5 g KCl 1 mole KCl 2 osmoles = 3.07 2 osmoles 74.55 g KCl 1 mole KCl Osmolarity = osmoles = 3.07 2 osmoles = 0.85 osmoles /L solution = 0.85 osmolar L of solution 3.6 L of solution c) How many osmoles are contained in 1.00 L of the above potassium chloride solution? 1.00 L solution 0.85 osmoles = 0.85 osmoles L of solution
  • As in the case of molarity (M) and molality ( m ) , the concentration (os molarity this time) gives us the relationship between the amount of solute and the amount of solution ….we use the concentration as a conversion factor !!!! d) How many liters (L) of this potassium chloride solution would contain 0.350 osmoles? 0.35 0 osmoles L of solution = 0. 41 L of solution 0.85 osmoles 5 ) 7.58 g of 2-­‐propanol (C 3 H 8 O) is added to enough water to make 1.50 L of solution. a) How many osmoles are in one mole of 2-­‐propanol when it dissolves? one mole of 2 -­‐propanol = one osmole (2-­‐propanol does not dissociate into ions) b) What is the osmolarity of the solution? Molar mass of 2 -­‐propanol = 60. 11 g/mole
  • Osmoles in solution: 7.58 g 2 -­‐propanol 1 mole 2 -­‐propanol 1 osmole = 0.126 osmoles
  1. 11 g 2 -­‐propanol 1 mole 2 -­‐propanol Osmolarity = osmoles = 0.126 osmoles = 0. 0840 osmoles /L solution L of solution 1.50 L of solution = 0.0840 osmolar Concentration of the solution

b) What volume (mL) of this solution would contain 0.0735 grams of glucose?

  • Use the concentration as a conversion factor! 0.0735 g glucose 100. mL = 2.94 mL of solution 2.50 g glucose c) How many grams of glucose would be present in 185 mL of this solution?
  • Use the concentration as a conversion factor! 185 mL solution 2.50 g glucose = 4.63 g glucose 100. mL solution
  1. 234.5 g of KCl is dissolved in enough water to make 3.6 L of solution. a) How many equivalents of potassium (K+) are in one mole of KCl when it dissolves? one mole of KCl = 1 Eq K+^ (recall that an equivalent is a mole of charge)
  • This relationship can be used as a conversion factor to convert between moles and equivalents : 1 Eq K+^ or 1 mole KCl 1 mole KCl 1 Eq K+ b) What is the concentration from potassium in (Eq K+/L)?
  • First get the moles of KCl then convert equivalents (Eq) : Molar mass of KCl = 74.55 g/mole
  • Equivalents (Eq) in solution : 234.5 g KCl 1 mole KCl 1 Eq K+^ = 3.146 Eq K+ 74.55 g KCl 1 mole KCl (Eq/L) = # Eq K+^ = 3.146 Eq K+^ = 0.8 7 Eq K+/L solution L of solution 3.6 L of solution Note: 2.50 % (w/v) means there are 2.50 g in 100 mL of solution = your conversion factor.

c) How many equivalents Eq of K+^ are contained in 0.700 L of the above potassium chloride solution?

  • As in the case of molarity (M), the concentration ( Eq/L this time) gives us the relationship between the amount of solute and the amount of solution ….we use the concentration as a conversion factor !!!! 0.700 L solution 0.87 Eq K+^ = 0. 61 Eq K+ L of solution d) How many liters (L) of this potassium chloride solution would contain 0. 050 equivalents Eq of K+^?
    1. 050 Eq K+^1 L of solution = 0. 057 L of solution 0.87 Eq K+ 9 ) 0.250 g of aluminum sulfate is dissolved in enough water to make 150 mL of solution. a) How many equivalents of sulfate ion (SO 42 -­‐) are in one mole of aluminum sulfate when it dissolves? one mole of Al 2 (SO 4 ) 3 = 6 Eq SO 42 -­‐^ (recall that an equivalent is a mole of charge/mole of compound) o 3 moles sulfate ions x (2 moles of charge/1 mole sulfate ions) = 6 Eq
  • This relationship can be used as a conversion factor to convert between moles and equivalents : 6 Eq SO 42 -­‐^ or 1 mole Al 2 (SO 4 ) 3 1 mole Al 2 (SO 4 ) 3 6 Eq SO 4 2 -­‐ b) What is the concentration of sulfate in (Eq/L)?
  • First get the moles of Al 2 (SO 4 ) 3 then convert equivalents (Eq) : Molar mass of Al 2 (SO 4 ) 3 = 342 .17 g/mole
  • Equivalents (Eq) in solution : 0.250 g Al 2 (SO 4 ) 1 mole Al 2 (SO 4 ) 3 6 Eq SO 42 -­‐^ = 0.00438 Eq SO 42 -­‐ 342.17 g Al 2 (SO 4 ) 3 1 mole Al 2 (SO 4 ) 3 o Note: we must convert from mL of solution to L of solution (Eq/L) = # Eq SO 42 -­‐^ = 0.00438 Eq SO 42 -­‐^ = 0. 029 Eq SO 42 -­‐/L solution L of solution 0.15 L of solution Concentration of potassium ions in solution

Molarity calculations (fill-­‐in all the boxes)

solute moles of

solute

grams of

solute

volume of

solution

Concentration

(Molarity, M=mole/L)

NaCl 3.00 moles 175 g 0.500 L 6.00 M

NaCl .231 moles 13.5 g .150 L 1.54 M

NaCl .375 moles 21.9 g .375 L 1.00 M

NaCl .0010 moles .059 g .0033 L 0.30 M

KNO 3 1.57 moles 159 g 2.04 L .770 M

KNO 3 .0196 moles 1.98 g .00980 L 2.00 M

KNO 3 .0567 moles 5.73 g .288 L .197 M

Osmolarity calculations

solute moles of

solute

osmoles of

solute

grams of

solute

volume of

solution

Concentration

(Osmolar = Osmole/L )

KCl 2.40 moles 4.8 0 osmoles 179 g 0.600 L 8.00 osmolar

KCl 0.020 moles 0.040 osmoles 1.5 g 0.750 L 0.053 osmolar

KCl .050 moles 0.10 osmoles 3.7 g 0.10 L 1.00 osmolar

KCl 0.0120 moles 0.0240 osmoles 0 .892 g 0.160 L 0.150 osmolar

glucose

C 6 H 12 O 6

1.50 moles 1.50 osmoles 270. g 1.23 L 1.22 osmolar

glucose

C 6 H 12 O 6

0.00649 moles 0.00649 osmoles 1.17 g .649 L 0.0100 osmolar

glucose

C 6 H 12 O 6

0.0281 moles 0.0281 osmoles 5.06 g 0.375 L 0.0750 osmolar