Question

1. What is the mass of 32 x 10²³ nitrogen gas particles?

2. If 15.2 pg of water is dissociated from lithium carbonate pentahydrate, what amount of lithium carbonate pentahydrate was originally used?

3. When 18.2 hg of water is used in a reaction, how many molecules were reacted? (assume that the reaction went to competion)

4. How many pg hydrogen are in 3.39 x 10¹⁵ ng of barium hydrogen carbonate hexahydrate?

5. A sample of lead contains 8.88 x 10¹⁵ micrograms, How many atoms does this sample contain?

6. Strontium nitrate consist of 72.8 mmol. How many pg of substance is this?

7. A sample of oxygen gas weighs 28.4 g. How many atoms of oxygen are present in this sample?

8. If you have one mole of ammonia, how many moles of hydrogen do you have?

9. Suppose you have half a mole of sulfur atoms. How many sulfur atoms do you have?

10. How many micrograms of zinc carbonate are there in 52.0 x 10^-5 moles?

Answer 1

Ans. #1. Moles of N2 gas = No. of particles / Avogadro number

                                                = 32 x 10²³ particles (6.022 x 10²³ particles/ mol)

                                                = 5.314 mol

Now,

            Mass of N₂ sample = Moles of N₂ x Molar mass

                                                = 5.314 mol x (28.01348 g/ mol)

                                                = 148.860 g

#2. Given, Mass of water dissociated = 15.2 pg = 15.2 x 10^-12 g

Moles of H₂O = Mass / Molar mass

= (15.2 x 10^-12 g) / (18.01528 g/mol)

= 8.4373 x 10^-13 mol

Dissociation of Li₂CO₃.5H₂O:                      Li₂CO₃.5H₂O- ---------> Li₂CO₃ + 5 H₂O

1 mol Li₂CO₃.5H₂O produces 5 mol H2O.

So,

            Moles of Li₂CO₃.5H₂O in sample = (1/5) x moles of H₂O produced

                                    = (1/5) x 8.4373 x 10^-13 mol

                                                            = 1.68745 x 10^-13 mol

            Moles of Li₂CO₃.5H₂O in sample = Moles x Molar mass

                                                            = 1.68745 x 10^-13 mol x (163.9676 g/mol)

                                                            = 2.767 x 10^-11 g

                                                            = 27.67 pg

#3. Mass of water = 18.2 hectogram

                                    = 18.2 x (100.0 g)                                         ; [1 hg = 100 g]

                                    = 1820.0 g

Moles of H₂O = 1820.0 g / (18.01528 g/mol) = 32787.8096 mol

Number of H₂O molecules = Moles x Avogadro number

                                                = 32787.8096 mol x (6.022 x 10²³ molecules / mol)

                                                = 1.975 x 10²⁸ molecules

#4. Mass of Ba(HCO₃)₂.6H₂O = 3.39 x 10¹⁵ ng = 3.39 x 10¹⁵ (10^-9 g) = 3.39 x 10⁶ g

Moles of Ba(HCO₃)₂.6H₂O = 3.39 x 10⁶ g / (367.45296 g/mol) = 9.2257 x 10³ mol

# 1 mol Ba(HCO₃)₂.6H₂O consists of 14 mol H-atom

So,

            Moles of H-atoms = 14 x moles of Ba(HCO₃)₂.6H2O

                                                = 14 x 9.2257 x 10³ mol

                                                = 1.2916 x 10⁵ mol

            Total mass of H-atoms = 1.2916 x 10⁵ mol x (1.00 g/mol) = 1.2916 x 10⁵ g

                                                = 1.2916 x 10⁵ x (10¹² pg)

                                                = 1.2916 x 10¹⁷ pg

#5. Mass of Pb = 8.88 x 10¹⁵ ug = 8.88 x 10⁹ g

Moles of Pb = 8.88 x 10⁹ g / (207.2 g/mol) = 4.2857 x 10⁷ mol

Number of Pb atoms = 4.2857 x 10⁷ mol x (6.022 x 10²³ atoms /mol)

= 2.5809 x 10³¹ atoms

#6. Moles of Sr(NO₃)₂ = 72.8 mmol = 0.0728 mol

Mass of Sr(NO₃)₂ sample = 0.0728 mol x (211.62988 g/mol)

                                                = 15.41 g

                                                = 1.541 x 10¹³ pg

#10. Given, moles of ZnCO₃ = 52.0 x 10^-5 mol

            Mass = (52.0 x 10^-5 mol) x (125.3992 g/mol) = 0.0652 g

                                                = 0.0652 x (10^-6 ug)

                                                = 6.52 x 10⁴ ug

Note: Please do 7. 8 and 9 on your own as time runs out. Let me know if you need any assistance.