Chemistry 11 Assignment 2 Total 25 marks
[pic 1]
Please make sure you highlight your final answer. Failure to do so will result in one mark taken off.
Part 1
For the next questions, clearly show all the steps in solving the problems. 0.5 marks will be deducted for each time there is a significant figure error or appropriate units are not used!
1. What is the molar mass (to the nearest tenth of a gram/mole) of each of the following (2 marks)
a. KCl
K: 1 x 39.1 = 39.1
Cl: 1 x 35.5 = 35.5
39.1 + 35.5 = 74.6 g/mol
b. Al2(SO4)3
Al: 2 x 27.0 = 54.0
S: 3 x 32.1 = 96.3
O: 12 x 16.0 = 192
96.3 + 54.0 + 192 = 342g/mol
2. Convert the following mass into moles (1 mark)
19.5 grams of Al(OH)3
Al: 1 x 27.0 = 27.0
O: 3 x 16.0 = 48.0
H: 3 x 1.0 = 3.0
27.0 + 48.0 + 3.0 = 78.0
19.5 grams Al(OH)3 x 1 mole / 78.0 grams Al(OH)3 = 0.250 mol Al(OH)3
3. Convert the following moles into masses (1 mark)
2.00 moles of N2
2.00 moles x 28.0 g / 1 mole =56.0g N2
4. How many molecules are present in 1.5 moles of NaCl? (1 mark)
1.5 moles NaCl x 6.02 x 10 ^23 molec NaCl/ 1 mole NaCl = 9.0 x 10^23 molec NaCl
5. How many atoms of chlorine are in 19.0 grams of MgCl2? (1 mark)
Mg: 1 x 24.31 = 24.31
Cl: 2 x 35.45 = 70.9
24.31 + 70.9 = 95.21 95.2 g/mol[pic 2]
19.0 g MgCl2 x 1 mol MgCl2 /95.2 g Mg Cl2 x 6.02 x 10^23 molec/ 1 mol x 2 atoms/ 1 molec = 2.40 x 10 ^23 atoms
6. What is the volume of 2.60 moles of CO2(g) at STP? (1 mark)
2.60 mol CO2 x 22.4 L CO2 / 1 mol CO2 = 58.24 L CO2 58.2 L CO2[pic 3]
7. What is the mass of 50.0 L of O2(g) at STP? (1 mark)
50.0 L O2 x 1 mol O2 / 22.4 L O2 x 15.99 g O2 / 1 mol O2 =35.7 g O2
8. What is the percent composition of chlorine in PCl5? (1 mark)
Cl: 5 x 35.45= 177.25
P: 1 x 30.97 = 30.97
177.25 + 30.97 = 208.22
Cl: 177.25/208.22 x 100% = 85.13 %
The percent composition of Chlorine is 85.13 %
9. Steven analyzed a compound and found it to be 52.7% Potassium and 47.3% Chlorine. He determined the empirical formula to be … (1 mark)
52.7 g K x 1 mole K / 39.1 g k = 1.35 mol K/ 1.33 ≈ 1
47.3 g Cl x 1 mole Cl/ 35.45 g Cl = 1.33 mol Cl/1.33 = 1
Empirical formula: KCl
10. If the empirical formula of a compound is CH2 and its molar mass is 70.0, then what is its molecular formula? (1 mark)
12.0 C + 2 x 1.0 H = 14.0g/mol
70.0 g/mol ÷ 14.0g/mol = 5
5 x CH2 = C5H10
11. If 0.30 mol NaCl goes into a solution with 4.00L water, the molarity of NaCl is (2 marks)
M = n / V
M= 0.30 mol / 4.00L
M= 0.0750 mol/ L
12. If 100.0 ml water is added to 150.0 ml of a 2.00M HCl solution, the new molarity of HCl is (2 marks)
150.0 mL x 1 L/ 1000 mL = 0.1500L
100.0Ml x 1L/1000mL= 0.1000L
n = mv