Unit 9 Final Lab Report

1. If atomic mass of Mg atom is 24 g, find mass of 1 Mg atom.
2. Find mass of 1 molecule C2H6. (C=12, H=1)
3. Find mole of 6,9 g Na. (Na=23)
4. Find mass of 0,2 mol P4 . (P=31)
5. Find mole of 4,48 liters O2 under normal conditions.
6. Find mass of Fe in the compound including 4,8x1023 O atoms ;Fe3O4 .
7. 1. Find relation between number of molecules of given matters;
8. I. C2H2 that includes 2mol H atom
9. II. CH4 that includes N atoms (N is Avogadro number)
10. III. C3H4 that includes 1,5 N C atoms
11. Find relation between number of atoms of given matters.
12. I. 6 PH3 molecules
13. II. CO2 that includes 24N atom
14. III. 8mol O3
15. If, 8,4 g X element includes 9,03x1022 atoms and 0,1mol X2Y3 compound is 16g find atomic mass of Y element. (Avogadro number is 6,02x1023)
16.
Which ones of the following statements are true for 0,2mol C3H4 and 0,5mol C2H6gas mixture? (C=12, H=1 and N=Avogadro Number)
17. Q. a sample of mixture of NaCl and NaBr weighing 0.180g is treated with AgNO_3 solution to give 0.3715g of precipitate. Calculate the content of NaCl and NaBr in the mixture.
18. Solution Both NaCl and NaBr give white and pale yellow ppts of AgCl and AgBr respectively on reaction with AgNO3.
Let the amount of NaCl in the mixture = y grams
so,
Let the amount of NaBr in the mixture = (0.180 - y) grams so,
Moles of NaCl = y/58.5
Moles of NaBr = (0.180 - y) / 103
1 mole NaCl gives 1 mole AgCl. So, y/58.5 moles NaCl will give y/58.5 moles AgCl
1 mole NaBr gives 1 mole AgBr. So, (0.180 - y)/103 moles NaCl will give (0.180 - y)/103 moles AgBr
Mass of AgCl = (y/58.5) x 143.5 g = 2.452 y
Mass of AgBr = (0.180 - y)/103 x 188 g = 1.825 (0.180 - y)
Now, 2.452y + 1.825 (0.180 - y) = 0.3715
So, 2.452y + 0.3285 - 1.825y = 0.3715
So, 0.627y = 0.043 or y = 0.068 g ie mass of NaCl
0.180 - 0.068 = 0.112 g ie mass of NaBr