General Chemistry Answers

1. State the effect that dissolving a solute has on each of the following physical properties of a solvent.

a. Boiling Point

b. Freezing Point

2. Explain why the colligative properties of a solvent are affected more by the dissolving of an electrolyte compared to an equal amount of a nonelectrolyte.

3. Are the values of Kf and Kb dependent on the identity of the solvent, the solute, or both? Explain.

Congratulations! You won the lotto and bought a new car. Since it's very hot in the philippines you are required to buy an antifreeze to keep the water in your car's radiator from boiling. You add 1.00kg of ethylene glycol (C2H6O2) antifreeze to 4450 g of water in your car's radiator. What are the boiling and freezing points of the solution?

1. Calculate the freezing and boiling points of a solution prepared by dissolving 15.5g of AI(NO3)3 in 200.0g of water. (Molar mass of AI(NO3)3 is 212.996 g/mol).

2. A solution is prepared by dissolving 120 grams of NaCI in 450 grams of water. Find the freezing and boiling point of this solution. (Molar mass of NaCI is 58.44 g/mol).

A solution was produced by dissolving 3.75 g of a nonvolatile solute in 95g of acetone. The boiling of pure acetone was observed to be 55.95 °C, while the boiling point of the solution was 56.50°C. If the boiling point for acetone is 1.71°C/mz what is the approximate molar mass of solute?

Step 1: First Compute the molality of the boiling point equation

Step 2: Then, from the definition of molality, compute the number of moles solute, n (solute, in the sample).

Step 3: Solve for the molar mass

A solution containg 4.50 g of a non electrolyte (i=1) dissolved in 125g of water freezes at -0.372°C. Calculate the molar mass of the solute.

Step 1: Firsy compute the molarity of the freezing point equation

Step 2: Then, from the definition of molality, compute the number of moles solute, n(solute), in the sample.

Step 3: Solve for the molar mass

The molal boiling point constant of water is 0.51°C/m. If 1 mol cane sugar (342g/mol) is dissolved in 1kg water, the solution will boil at 100.512°C, assuming standard pressure. By this relationship, a half of a module of sugar (171g) would boil at 100 256 C/m and 2 mol sugar (684g) should boil at 101.024°C/m. However, these are not necessarily solutions that can be mixed; 684g sugar of sugar is nearly a pound and a half of sugar, which you are supposed to dissolved in a liter of water!

From the given situation above, why is it important to underatand the constant relating to water?

A 15.00 g sample containing mixed alkali and other inert components was dissolved and diluted to 300 mL with water. A 20 mL aliquot was titrated with 5.02 mL of 0.5352 M HCl to reach PHP endpoint. Another 20 mL aliquot was titrated to the BCG endpoint, using up 18.87 mL of titrant in the process. How much volume of titrant was needed to reach the second endpoint starting from the first endpoint? Answer in two decimal places.

1. The molal freezing pointof water is 1.86°C/m(1.86K/m), which can be expressed as 1.86°C • kg H2O/mol solute) after some algebraic rearrangement. This means that a solution of 1 mol cane sugar (342g, over 3/4 lb) dissolvedin 1kgof water should freeze at -1.86°C.

Describe the freezing point drop suggestin number 1.

Pb(C2H3O2)2(aq) + (NH4)2SO4(aq) → Oxidation # of sulfur?

3Pb(C2H3O2)4(aq) + 4K3PO4(aq) → Pb3(PO4)4(s) + 12KC2H3O2(aq) What's wrong with the reaction?