Calculations for Solutions Worksheet and Key
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1 Calculations for Solutions Worksheet and Key 1) 23.5g of NaCl is dissolved in enough water to make.683. a) What is the molarity (M) of the solution? b) How many moles of NaCl are contained in L of the above NaCl solution? c) What volume (L) of this NaCl solution would contain moles of NaCl? 2) 12.5g of glucose (C 6 H 12 O 6 ) is dissolved in enough water to make m. 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 moles of glucose?
2 3) 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 moles of MgCl 2? 4) g of KCl is dissolved in enough water to make 3.6. 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 osmoles?
3 5) 7.58 g of propanol (C 3 H 8 O) is added to enough water to make a) How many osmoles are in one mole of propanol when it dissolves? b) What is the osmolarity of the solution? c) How many osmoles are contained in ml of the above propanol solution? d) How many liters (L) of this propanol solution would contain osmoles? 6) 46.0 g of barium nitrate is dissolved in 2.60 kg of water. a) How many osmoles are in one mole of barium nitrate when it dissolves? b) What is the osmolality of the solution? 7) A glucose (C 6 H 12 O 6 ) solution is prepared by adding 5.00 grams of glucose to enough water to make ml of solution. a) What is the %(w/v) of the solution? b) What volume (ml) of this solution would contain grams of glucose? c) How many grams of glucose would be present in 185 ml of this solution?
4 8) g of KCl is dissolved in enough water to make 3.6. 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 L of the above potassium chloride solution? d) How many liters (L) of this potassium chloride solution would contain equivalents Eq of K +? 9) g of aluminum sulfate is dissolved in enough water to make 150 m. a) How many equivalents of sulfate ion (SO 4 ) are in one mole of aluminum sulfate when it dissolves? (note: you are concerned with the Eq from SO 4 only, do not include Eq from Al + ) b) What is the concentration of sulfate in (Eq/L)? c) How many equivalents (Eq) of SO 4 are contained in L of the above aluminum sulfate solution? d) How many liters (L) of this aluminum sulfate solution would contain equivalents Eq of SO 4?
5 Molarity calculations (fill- in all the boxes) moles of grams of volume of solution NaCl 3.00 moles L Concentration (Molarity, M=mole/L) NaCl 13.5 g L NaCl moles 1.00 M NaCl g 0.30 M KNO moles M KNO g 2.00 M KNO L M Osmolarity calculations moles of osmoles of grams of volume of solution KCl 2.40 moles L Concentration (Osmolar = osmole/l) KCl 1.5 g L KCl moles 1.00 osmolar KCl g osmolar glucose 1.50 moles 1.22 osmolar C 6 H 12 O 6 glucose 1.17 g osmolar C 6 H 12 O 6 glucose C 6 H 12 O L osmolar
6 Key 1) 23.5g of NaCl is dissolved in enough water to make.683. a) What is the molarity (M) of the solution? Molar mass of NaCl = g/mole Moles of NaCl: 23.5 g NaCl 1 mole NaCl =.402 moles NaCl gnacl Molarity = moles = moles NaCl = moles NaCl/L = M NaCl liter solution b) How many moles of NaCl are contained in L of the above NaCl solution? Concentration of the solution L solution moles NaCl = mole NaCl Note: The concentration gives us the relationship between the amount of and the amount of solution.we use the concentration as a conversion factor!!!! c) What volume (L) of this NaCl solution would contain moles of NaCl? Concentration of the solution moles of NaCl = moles NaCl 2) 12.5g of glucose (C 6 H 12 O 6 ) is dissolved in enough water to make m. a) What is the molarity (M) of the solution? Molar mass of glucose = g/mole Moles of glucose: 12.5 g glucose 1 mole glucose = moles glucose g glucose Molarity = moles = moles glucose = mole glucose/l = M glucose liter solution b) How many moles of glucose are contained in 237 ml of the above glucose solution? L solution moles glucose = mole glucose
7 c) What volume (L) of this glucose solution would contain moles of glucose? moles glucose = moles glucose 3) 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 = g/mole Moles of MgCl 2 : 45.7 g MgCl 2 1 mole MgCl 2 = moles MgCl g MgCl 2 Molality = moles = moles MgCl 2 = moles MgCl 2 /kg = 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? Concentration of the solution 1.76 kg solvent moles MgCl 2 = moles MgCl 2 1 kg of solvent c) How many kg of solvent would contain moles of MgCl 2? Concentration of the solution moles MgCl 2 1 kg of solvent = kg of solvent moles MgCl 2 4) g of KCl is dissolved in enough water to make 3.6. 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
8 b) What is the osmolarity of the solution? First get the moles of KCl then convert to osmoles: Molar mass of KCl = g/mole Osmoles in solution: g KCl 1 mole KCl 2 osmoles = osmoles g KCl 1 mole KCl Osmolarity = osmoles = osmoles = 0.85 osmoles /L solution = 0.85 osmolar 3.6 c) How many osmoles are contained in 1.00 L of the above potassium chloride solution? Concentration of the solution 1.00 L solution 0.85 osmoles = 0.85 osmoles As in the case of molarity (M) and molality (m), the concentration (osmolarity this time) gives us the relationship between the amount of 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 osmoles? osmoles = osmoles 5) 7.58 g of propanol (C 3 H 8 O) is added to enough water to make a) How many osmoles are in one mole of propanol when it dissolves? one mole of propanol = one osmole (propanol does not dissociate into ions) b) What is the osmolarity of the solution? Molar mass of propanol = g/mole Osmoles in solution: 7.58 g propanol 1 mole propanol 1 osmole = osmoles g propanol 1 mole propanol Osmolarity = osmoles = osmoles = osmoles /L solution 1.50 = osmolar
9 c) How many osmoles are contained in ml of the above propanol solution? L solution osmoles = osmoles d) How many liters (L) of this propanol solution would contain osmoles? osmoles = osmoles 6) 46.0 g of barium nitrate is dissolved in 2.60 kg of water. a) How many osmoles are in one mole of barium nitrate when it dissolves? one mole of Ba(NO 3) 2 = 3 osmoles Ba(NO 3) 2 dissociates into 3 particles, one Ba 2+ ion and 2 nitrate ions This relationship can be used as a conversion factor to convert between moles and osmoles: 3 osmoles or 1 mole Ba(NO 3 ) 2 1 mole Ba(NO 3 ) 2 3 osmoles b) What is the osmolality of the solution? Molar mass of Ba(NO 3) 2 = g/mole Osmoles in solution : 46.0 g Ba(NO 3 ) 2 1 mole Ba(NO 3 ) 2 3 osmoles = osmoles Ba(NO 3 ) g Ba(NO 3 ) 2 1 mole Ba(NO 3 ) 2 Osmolality = osmoles = osmoles = osmoles/kg kg of solvent 2.60 kg of solvent = osmolal 7) A glucose (C 6 H 12 O 6 ) solution is prepared by adding 5.00 grams of glucose to enough water to make ml of solution. a) What is the %(w/v) of the solution? %(w/v) = g x 100 = 5.00 g glucose x 100% = 2.50 % (w/v) m ml
10 b) What volume (ml) of this solution would contain grams of glucose? Use the concentration as a conversion factor! g glucose 100. ml = 2.94 m 2.50 g glucose Note: 2.50 % (w/v) means there are 2.50 g in 100 ml of solution = your conversion factor. 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 8) g of KCl is dissolved in enough water to make 3.6. 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 = g/mole Equivalents (Eq) in solution : g KCl 1 mole KCl 1 Eq K + = Eq K g KCl 1 mole KCl (Eq/L) = # Eq K + = Eq K + = 0.87 Eq K + /L solution 3.6
11 c) How many equivalents Eq of K + are contained in 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 and the amount of solution.we use the concentration as a conversion factor!!!! Concentration of potassium ions in solution L solution 0.87 Eq K + = 0.61 Eq K + d) How many liters (L) of this potassium chloride solution would contain equivalents Eq of K +? Eq K + 1 = Eq K + 9) g of aluminum sulfate is dissolved in enough water to make 150 m. a) How many equivalents of sulfate ion (SO 4 ) are in one mole of aluminum sulfate when it dissolves? one mole of Al 2 (SO 4 ) 3 = 6 Eq SO 4 (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 4 or 1 mole Al 2 (SO 4 ) 3 1 mole Al 2 (SO 4 ) 3 6 Eq SO 4 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 = g/mole Equivalents (Eq) in solution : g Al 2 (SO 4 ) 1 mole Al 2 (SO 4 ) 3 6 Eq SO g Al 2 (SO 4 ) 3 1 mole Al 2 (SO 4 ) 3 = Eq SO 4 o Note: we must convert from m to (Eq/L) = # Eq SO 4 = Eq SO = Eq SO 4 /L solution
12 c) How many equivalents (Eq) of SO 4 are contained in L of the above aluminum sulfate solution? As in the case of molarity (M), the concentration (Eq/L in this case) gives us the relationship between the amount of and the amount of solution.we use the concentration as a conversion factor!!!! Concentration of sulfate ions in solution L solution Eq SO 4 = Eq SO 4 d) How many liters (L) of this aluminum sulfate solution would contain equivalents Eq of SO 4? Eq SO 4 1 = Eq SO 4
13 Molarity calculations (fill- in all the boxes) moles of grams of volume of solution Concentration (Molarity, M=mole/L) NaCl 3.00 moles 175 g 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 moles 159 g 2.04 L.770 M KNO moles 1.98 g L 2.00 M KNO moles 5.73 g.288 L.197 M Osmolarity calculations moles of osmoles of grams of volume of Concentration solution (Osmolar = Osmole/L) KCl 2.40 moles 4.80 osmoles 179 g L 8.00 osmolar KCl moles osmoles 1.5 g L osmolar KCl.050 moles 0.10 osmoles 3.7 g 0.10 L 1.00 osmolar KCl moles osmoles g L osmolar glucose 1.50 moles 1.50 osmoles 270. g 1.23 L 1.22 osmolar C 6 H 12 O 6 glucose moles osmoles 1.17 g.649 L osmolar C 6 H 12 O 6 glucose C 6 H 12 O moles osmoles 5.06 g L osmolar
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