Additional Chapter 7 Homework Problems: Due with chapter 7 homework, show your work for full credit!

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1 Additional Chapter 7 Homework Problems: Due with chapter 7 homework, show your work for full credit! Note: If you are struggling with these, see the chapter 7 worksheet titled: Molarity, Molality, Osmolality, Osmolarity Worksheet, where you can see the complete solutions for similar problems. 1) 5.75 g of LiCl is dissolved in enough water to make L of solution. a) What is the molarity (M) of the solution? Answer: M LiCl b) How many moles of LiCl are contained in L of the above LiCl solution? Answer: mole LiCl Note: The concentration is the relationship between the amount of solute and the amount of solution.we use the concentration as a conversion factor!!!! c) What volume (L) of this LiCl solution would contain moles of LiCl? Answer: L of solution

2 2) 1.25g of acetone (C 3 H 6 O) is dissolved in enough water to make ml of solution. a) What is the molarity (M) of the solution? Answer: mole acetone /L = M b) How many moles of acetone are contained in 679 ml of the above acetone solution? Answer: mole acetone c) What volume (L) of this acetone solution would contain moles of acetone? Answer: 2.0 L of solution

3 3) g of magnesium chloride is dissolved in 1.37 kg of water. a) Write the correct formula for magnesium chloride. b) What is the molality (m) of the solution? 4) 11.5 g of NaCl is dissolved in enough water to make 5.30 L of solution. Answer: = moles MgCl 2 /kg = m MgCl 2 a) How many osmoles are in one mole of NaCl when it is dissolved? (check your notes for the definition of osmoles) Answer: one mole of NaCl = 2 osmoles (1 mole of Na + particles + 1 mole of Cl - particles = 2 moles of dissolved particles) b) What is the osmolarity of the solution? Answer: = osmolar

4 c) How many osmoles of sodium chloride are contained in 2.00 L of the above sodium chloride solution? Answer: osmoles As in the case of molarity (M) and molality (m), the concentration (osmolarity 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 sodium chloride solution would contain 3.50 osmole? Answer: 47.1 L of solution 5) 1.38 g of 2-propanol (C 3 H 8 O) is added to enough water to make 2.25 L of solution. a) How many osmoles are in one mole of 2-propanol when it is solvated? (Hint:2-propanol is solvated by water, but does not dissociate into ions) b) What is the osmolarity of the solution? Answer: osmoles /L solution = osmolar

5 c) How many osmoles are contained in ml of the above 2-propanol solution? Answer: osmoles d) How many liters (L) of this 2-propanol solution would contain 2.00 osmoles? Answer: 196 L of solution 6) 1.05 g of Magnesium nitrate is dissolved in 2.88 kg of water. a) How many osmoles of magnesium nitrate are in one mole of maginesium nitrate when it is dissolved? Hint: Mg(NO 3 ) 2 dissociates into 3 particles, one Mg 2+ ion and 2 nitrate ions b) What is the osmolality of the solution? Answer: osmoles = osmolal

6 7) A glucose solution is prepared by adding 1.47 grams of glucose to enough water to make ml of solution. a) What is the %(w/v) of the solution? % (w/v) b) What volume (ml) of this solution would contain grams of glucose? c) How many grams of glucose would be present in 345 ml of this solution? 10.2 ml of solution 1.69 g glucose 8) If 45 g of glucose dissolved in g of water, what is the %(w/w) concentration? 18%(w/w)

7 9) g of KCl is dissolved in enough water to make 1.2 L of solution. a) How many equivalents of potassium (K + ) are in one mole of KCl when it dissolves? Answer: one mole of KCl = 1 Eq K + (recall that an equivalent is a mole of charge) b) What is the concentration of potassium in (Eq/L)? answer: 1.5 Eq K + /L solution c) How many equivalents (Eq) of K + are contained in L of the above potassium chloride solution? answer: 0.11 Eq K + d) How many liters (L) of this potassium chloride solution would contain 3.5 equivalents (Eq) of K +? answer = 2.3 L of solution

8 10) g of Iron(III) sulfate is dissolved in enough water to make 75 ml of solution. a) How many equivalents of sulfate ion (SO 4 2- ) are in one mole of iron(iii) sulfate when it dissolves? Answer: one mole of Fe 2 (SO 4 ) 3 = 6 Eq SO 4 2- (recall that an equivalent is a mole of charge/mole of compound) b) What is the concentration of sulfate in (Eq/L)? Answer: 0.10 Eq SO 4 2- /L solution c) How many equivalents (Eq) of SO 4 2- are contained in 7.80 L of the above iron(iii) sulfate solution? Answer: 0.78 Eq SO 4 2- d) How many liters (L) of this iron(iii) sulfate solution would contain 0.95 equivalents (Eq) of SO 4 2-? Answer: 9.5 L of solution