Part 1 Chapt 4
1. You dissolve 157.1 g of ammonium nitrate in a 250 mL volumetric flask and add water to the mark (call this solution A). You take 30 mL of solution A and place it in a 100 mL volumetric flask and add water to the mark (this is solution B). Next you take 50 mL of solution B and place it in a 500 mL volumetric flask and add water to the mark (this is solution C). Finally you take 10 mL of solution A, 10 mL of solution B and 10 mL of solution C and mix them to form solution D. What is the concentration (M) of solution D? (see problem 6 in Zumdahl test bank)
2. Zn(OH)2 solid (molecular mass = 99.41 amu) is added to 0.500 L of 0.50 M HBr. The resultant solution which is acidic, requires 185 mL of 0.500 M NaOH for neutralization. Calculate the mass (in mg) of Zn(OH)2 that was originally added (problem 53 in Purdue test bank)
3. A student made up a solution of 30 g of barium chloride in a 2 L volumetric flask. Then she placed 100 mL of this solution in a 500 mL beaker. Over the weekend, the roof in the lab leaked and some water ended up in the beaker. Then some water evaportated. After several days she analyzed it with silver nitrate to find it was now 0.25 M in chloride ion. How much water must have been added or evaporated from the beaker? (see problem 14 in Purdue test bank or 20 in Zumdahl test bank)
4. Calculate the concentration (M) of all the ions present in a solution when 20 g of sodium arsenate is dissolved in a 250 mL volumetric flask and diluted with water to the mark. When 0.1 M silver nitrate is added to 10 mL of the above solution how many grams of solid is precipitated and what is the solid? (see lab work)
5. The iron content of ores can be determined by titrating a sample with a solution of potassium permanganate (KMnO4). This is a redox reaction in which the ore is dissolved with hydrochloric acid to give Fe2+ ions that are titrated with MnO4-. For a particular ore sample, 0.202 g of dissolved ore was reacted with 20.1 mL of 0.0101 M potassium permanganate. How much iron (as Fe2+) was present in grams and %? (question 72 in test bank for Zumdahl )
Part 2 Chapt 5 Gas Laws.
6. A scientist has an empty long glass tube connected with stopcocks (valves) to three flasks. The volume of the tube is 1 L. The first flask contains 250 mL of hydrogen gas at 250 torr. The second flask has 1.5 g of argon in 1L and the third flask is 2L and contains 0.00 4 moles of oxygen. The temperature in the lab is 27oC. After the scientist opens all three stopcocks so all three flasks and the tube mix together, what will be the pressure inside the tube? (See text)
7. 10 L of ammonia and 9 L of chlorine are mixed at STP and the following reaction occurs:
Calculate the volume of each gas. (question 47 Zumdahl test bank).
8. Calcium metal reacts with hydrochloric acid to form aqueous calcium chloride and hydrogen gas. If we react excess calcium metal with 40 mL of 3.00 M hydrochloric acid and the hydrogen is collected in a balloon at 27oC and 0.97 atm, what will be the volume of the balloon? If we collect the hydrogen in an inverted bottle full of water at the same temperature what will be the volume. The partial pressure of water is 28 mm-Hg. (Zumdahl test bank question 49)
9. A sample of xenon gas has a volume of 300 mL at STP. What volume does the gas occupy if the number moles, absolute temperature and pressure are each doubled. (see Zumdahl test bank question 17).
10. A 100 L cylinder contains 40% He (g) by weight and 60% Kr (g) by weight at 33oC and 1.65 atm total pressure. What is the partial pressure of He in the cylinder? (Zumdahl test bank 53).
1. To calculate the concentration (M) of solution D, we need to find the number of moles of the solute (ammonium nitrate) in solution D and divide it by the volume of solution D.
First, let's find the number of moles of ammonium nitrate in solution A:
- Given mass of ammonium nitrate = 157.1 g
- Molar mass of ammonium nitrate (NH4NO3) = 80.05 g/mol
- Moles of ammonium nitrate in solution A = mass / molar mass = 157.1 g / 80.05 g/mol = 1.962 mol
Next, let's find the number of moles of ammonium nitrate in solution B:
- Volume of solution B = 30 mL = 0.030 L
- Moles of ammonium nitrate in solution B = (moles in solution A) * (volume of solution B / volume of solution A) = 1.962 mol * (0.030 L / 0.250 L) = 0.2354 mol
Now, let's find the number of moles of ammonium nitrate in solution C:
- Volume of solution C = 50 mL = 0.050 L
- Moles of ammonium nitrate in solution C = (moles in solution B) * (volume of solution C / volume of solution B) = 0.2354 mol * (0.050 L / 0.100 L) = 0.1177 mol
Finally, let's find the number of moles of ammonium nitrate in solution D:
- Volume of solution D = 10 mL = 0.010 L
- Moles of ammonium nitrate in solution D = (moles in solution A) * (volume of solution D / volume of solution A) + (moles in solution B) * (volume of solution D / volume of solution B) + (moles in solution C) * (volume of solution D / volume of solution C)
= 1.962 mol * (0.010 L / 0.250 L) + 0.2354 mol * (0.010 L / 0.100 L) + 0.1177 mol * (0.010 L / 0.050 L)
= 0.07848 mol + 0.02354 mol + 0.02354 mol = 0.12556 mol
Now, we can calculate the concentration (M) of solution D:
- Concentration (M) = moles of ammonium nitrate / volume of solution D = 0.12556 mol / 0.010 L = 12.556 M
Therefore, the concentration of solution D is 12.556 M.
(Note: The steps provided above explain how to obtain the answer, but performing these calculations in a lab setting requires accurate measurements and proper use of equipment.)
2. To calculate the mass of Zn(OH)2 originally added, we need to use the concept of stoichiometry and the volume of NaOH required for neutralization.
First, let's calculate the number of moles of NaOH used:
- Volume of NaOH used = 185 mL = 0.185 L
- Moles of NaOH used = concentration * volume = 0.500 M * 0.185 L = 0.0925 mol
Now, let's calculate the number of moles of HBr in the reaction:
- Moles of HBr = concentration * volume = 0.50 M * 0.500 L = 0.250 mol
Since the balanced equation is 2NaOH + HBr -> Na2O + 2H2O, we can determine the stoichiometry between NaOH and HBr.
- From the balanced equation, 2 moles of NaOH react with 1 mole of HBr.
Therefore, the number of moles of NaOH used is twice the number of moles of HBr used.
- Moles of NaOH used = 2 * Moles of HBr used = 2 * 0.250 mol = 0.500 mol
Next, let's calculate the number of moles of Zn(OH)2 that reacted with HBr:
- From the balanced equation, 2 moles of NaOH react with 1 mole of Zn(OH)2.
Therefore, the number of moles of Zn(OH)2 is the same as the number of moles of NaOH used.
- Moles of Zn(OH)2 = Moles of NaOH used = 0.500 mol
Finally, let's calculate the mass of Zn(OH)2:
- Mass of Zn(OH)2 = Moles of Zn(OH)2 * molar mass = 0.500 mol * 99.41 g/mol = 49.705 g
Therefore, the mass of Zn(OH)2 originally added is 49.705 g.