1
3. Which definition matches which chemical term?
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Avogadro’s constant
molar mass
amount of substance
mole
mass percentage
concentration
mass of 1 mol of a substance
unit for the amount of substance
number of particles in 1 mol of a substance
unit is mol/l
symbol is n
calculated by dividing mass of solute by total
mass of solution
B
4. Calculate.
d) What is the concentration of a solution containing 2,3 g of ammonium chloride, NH4Cl, in 500 ml of water?
𝑛 𝑁𝐻4 𝐶𝑙 =
𝑐=
2,3 𝑔
2,3 𝑔
=
𝑔 = 0,04299 𝑚𝑜𝑙
𝑀 𝑁𝐻4 𝐶𝑙
53,5
𝑚𝑜𝑙
𝑛 0,04299 𝑚𝑜𝑙
𝑚𝑜𝑙
=
= 0,08589
≈ 0,086 𝑚𝑜𝑙/𝑙
𝑉
0,5 𝑙
𝑙
e) Express the concentration of the following solution in mass percentage: 0,45 g of NaCl in 150 ml distilled water.
𝑚−%=
0,45 𝑔
= 0,002991 = 0,2991 % ≈ 0,30 %
0,45 𝑔 + 150 𝑔
f) How many milliliters of water must be added to 50 ml of 0,3 M silver nitrate solution, AgNO3 (aq), to obtain 0,05 M?
𝑐1 𝑉1 = 𝑐2 𝑉2
0,3
𝑚𝑜𝑙
𝑚𝑜𝑙
∙ 0,05𝑙 = 0,05
∙ 𝑉2
𝑙
𝑙
𝑉2 = 0,3 𝑙
𝑡𝑕𝑒𝑟𝑒𝑓𝑜𝑟𝑒 250 𝑚𝑙 𝑚𝑢𝑠𝑡 𝑏𝑒 𝑎𝑑𝑑𝑒𝑑 𝑠𝑜 𝑡𝑕𝑎𝑡 𝑡𝑕𝑒 𝑓𝑖𝑛𝑎𝑙 𝑣𝑜𝑙𝑢𝑚𝑒 𝑖𝑠 300 𝑚𝑙.
C
2
5. Calculate.
a)
You are to produce 500 ml of ca. 0,100 mol/l nitric acid solution. You are given 36 % nitric acid (mass
percentage), its density being 1,214 kg/l, distilled water, a burette, pipettes and volumetric flasks. How
will you proceed?
𝑛 = 𝑐𝑉 = 0,5 𝑙 ∙ 0,100
𝑚𝑜𝑙
= 0,05 𝑚𝑜𝑙 𝐻𝑁𝑂3 𝑖𝑠 𝑛𝑒𝑒𝑑𝑒𝑑 𝑓𝑜𝑟 𝑡𝑕𝑒 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛.
𝑙
𝑚 𝐻𝑁𝑂3 = 𝑛𝑀 = 0,05 𝑚𝑜𝑙 ∙ 63,02
𝑔
= 3,151 𝑔
𝑚𝑜𝑙
𝐵𝑢𝑡 𝑜𝑛𝑙𝑦 36% 𝑜𝑓 𝑡𝑕𝑒 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑡𝑕𝑒 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑒𝑑 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑖𝑠 𝐻𝑁𝑂3 . 𝑇𝑕𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝑡𝑕𝑒 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑡𝑕𝑒 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑒𝑑
𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑡𝑕𝑎𝑡 𝐶𝑂𝑁𝑇𝐴𝐼𝑁𝑆 3,151 𝑔 𝑜𝑓 𝐻𝑁𝑂3 𝑖𝑠
𝐴𝑠 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑖𝑠 𝜌 =
3,151 𝑔
= 8,753 𝑔
0,36
𝑚
𝑚
8,753 𝑔
𝑐𝑎𝑛 𝑡𝑕𝑒 𝑣𝑜𝑙𝑢𝑚𝑒 𝑏𝑒 𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒𝑑: 𝑉 = =
= 7,201 𝑚𝑙
𝑉
𝜌 1,214 𝑔/𝑚𝑙
𝑆𝑂: 7,2 𝑚𝑙 𝑜𝑓 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑒𝑑 𝑛𝑖𝑡𝑟𝑖𝑐 𝑎𝑐𝑖𝑑 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑖𝑠 𝑚𝑒𝑎𝑠𝑢𝑟𝑒𝑑 𝑤𝑖𝑡𝑕 𝑎 𝑏𝑢𝑟𝑒𝑡𝑡𝑒 𝑜𝑟 𝑔𝑟𝑎𝑑𝑢𝑎𝑡𝑒𝑑 𝑝𝑖𝑝𝑒𝑡𝑡𝑒
𝑖𝑛𝑡𝑜 𝑎 500 𝑚𝑙 𝑣𝑜𝑙𝑢𝑚𝑒𝑡𝑟𝑖𝑐 𝑓𝑙𝑎𝑠𝑘 𝑎𝑛𝑑 𝑓𝑖𝑙𝑙𝑒𝑑 𝑤𝑖𝑡𝑕 𝑤𝑎𝑡𝑒𝑟 𝑡𝑜 𝑡𝑕𝑒 𝑚𝑎𝑟𝑘.
c)
Iodine can be produced in aqueous sulfuric acid as followed:
NaI (aq) + H2SO4 (aq) + MnO2 (s)  Na2SO4 (aq) + MnSO4 (aq) + I2 (s) + H2O (l)
Balance the equation. How much iodine can be produced from 150 ml of 0,50 M aqueous sodium iodide
and 2,9 g manganese dioxide?
2 NaI (aq) + 2 H2SO4 (aq) + MnO2 (s)  Na2SO4 (aq) + MnSO4 (aq) + I2 (s) + 2 H2O (l)
𝑛 𝑁𝑎𝐼 = 𝑐𝑉 = 0,50
𝑛 𝑀𝑛𝑂2 =
𝑚𝑜𝑙
∙ 0,15 𝑙 = 0,075 𝑚𝑜𝑙
𝑙
𝑚
2,9 𝑔
=
𝑔 = 0,0334 𝑚𝑜𝑙
𝑀
86,94
𝑚𝑜𝑙
𝑇𝑤𝑖𝑐𝑒 𝑎𝑠 𝑚𝑢𝑐𝑕 𝑜𝑓 𝑁𝑎𝐼 𝑖𝑠 𝑛𝑒𝑒𝑑𝑒𝑑 𝑡𝑕𝑎𝑛 𝑜𝑓 𝑀𝑛𝑂4 . 0,075 𝑚𝑜𝑙 > 2 ∙ 0,0334 𝑠𝑜 𝑀𝑛𝑂2 𝑟𝑢𝑛𝑠 𝑜𝑢𝑡 𝑓𝑖𝑟𝑠𝑡.
𝑛 𝐼2 = 𝑛 𝑀𝑛𝑂2 = 0,0334 𝑚𝑜𝑙
𝑚 𝐼2 = 𝑛𝑀 = 8,5 𝑔