Question

1. A solution is made by mixing 28.8 ml of water with 7.73g of sodium chloride. What is the resulting concentration (in molarity) of sodium ions in the solution? (Be sure to take into account the volume change due to adding in the sodium chloride which has a density of 2.17g/ml) Your answer should have two decimals.

2. A mysterious metal is found and you decided to use a method similar to emission spectroscopy to determine the nuclear charge. From the experiment it is observed that the electron transition from n=7 to n=5 has an associated wavelength of 625nm. What is the nuclear charge (Z) of this mysterious material? NOTE: due to shielding your Z value may not be a whole number.  
Your answer should have two decimal places and do not use scientific notation.

3. The combustion of acetylene (as in acetylene torches, shown below) liberates the intense heat needed for welding metals together:
2 C₂H₂ (g) + 5 O₂ (g) --> 4 CO₂ (g) + 2 H₂O (g)         \DeltaΔH = -2511 kJ
acetylene
If 2660.9 kJ of heat is produced by the reaction, how much acetylene, in grams, must have reacted? Assume excess oxygen is present.Make sure you report the answer to one decimal place and do not use scientific notation.

Answer 1

1.

Molar mass of sodium chloride = 58.4 g/mol

Moles of sodium chloride = mass of sodium chloride / molar mass of sodium chloride

                                         = 7.73 g / 58.4 g/mol

                                         = 0.132 mol

Volume of sodium chloride = mass of sodium chloride/density of sodium chloride

                                           = 7.73 g / 2.17 g/mL

                                           = 3.56 mL

Total volume of the solution = volume of water + volume of sodium chloride

                                             = 28.8 mL + 3.56 mL

                                             = 32.4 mL

                                             = 0.0324 L                       ( 1 mL = 0.001 L)

Molarity = moles of solute / liter of solution

              = 0.132 mol / 0.0324 L

              = 4.07 M

Now, 1 M of sodium chloride solution gives 1 M of Na⁺ solution.

NaCl(aq) Na⁺(aq) + Cl^-(aq)

Hence, 4.07 M of sodium chloride solution will give 4.07 M of Na⁺ solution.

Therefore, the molarity of Na⁺ ion in the solution = 4.07 M