Concentration Calculations
Worksheet
Determining the Concentration
1. 40.0g of NaOH is dissolved in 250cm3 of water. What is the concentration in moldm-3?
Working Out
Mr(NaOH) = 23.0 + 16.0 + 1.0 = 40.0
n = 40.0 / 40.0 = 1.00 mol
V = 250 / 1000 = 0.250 dm3
C = 1.00 / 0.250 = 4.00 mol dm-3
Answer = 4.00 mol dm-3
n = 40.0 / 40.0 = 1.00 mol
V = 250 / 1000 = 0.250 dm3
C = 1.00 / 0.250 = 4.00 mol dm-3
Answer = 4.00 mol dm-3
2. 12.0g of anhydrous iron (II) sulfate is dissolved in 25cm3 of water. What is the concentration?
Working Out
Mr(FeSO4) = 55.8 + 32.1 + 64.0 = 151.9
n = 12.0 / 151.9 = 0.0790 mol
V = 25 / 1000 = 0.025 dm3
C = 0.0790 / 0.025 = 3.16 mol dm-3
(25 is 2 s.f.)
Answer = 3.2 mol dm-3
n = 12.0 / 151.9 = 0.0790 mol
V = 25 / 1000 = 0.025 dm3
C = 0.0790 / 0.025 = 3.16 mol dm-3
(25 is 2 s.f.)
Answer = 3.2 mol dm-3
3. 6.00g of sodium hydrogencarbonate is dissolved in 20cm3 of water. What is the concentration?
Working Out
Mr(NaHCO3) = 23.0 + 1.0 + 12.0 + 48.0 = 84.0
n = 6.00 / 84.0 = 0.0714 mol
V = 20 / 1000 = 0.020 dm3
C = 0.0714 / 0.020 = 3.57 mol dm-3
(20 is 2 s.f.)
Answer = 3.6 mol dm-3
n = 6.00 / 84.0 = 0.0714 mol
V = 20 / 1000 = 0.020 dm3
C = 0.0714 / 0.020 = 3.57 mol dm-3
(20 is 2 s.f.)
Answer = 3.6 mol dm-3
4. 46.0g of copper sulfate hexahydrate is dissolved in 200cm3 of water. What is the concentration?
Working Out
(Note: Standard copper sulfate is pentahydrate, but hexahydrate (6H2O) specified)
Mr(CuSO4·6H2O) = 159.6 + 6(18.0) = 267.6
n = 46.0 / 267.6 = 0.172 mol
V = 0.200 dm3
C = 0.172 / 0.200 = 0.860 mol dm-3
Answer = 0.860 mol dm-3
Mr(CuSO4·6H2O) = 159.6 + 6(18.0) = 267.6
n = 46.0 / 267.6 = 0.172 mol
V = 0.200 dm3
C = 0.172 / 0.200 = 0.860 mol dm-3
Answer = 0.860 mol dm-3
5. 250g of potassium manganate (VII) is dissolved in 750cm3 of water. What is the concentration?
Working Out
Mr(KMnO4) = 39.1 + 54.9 + 64.0 = 158.0
n = 250 / 158.0 = 1.58 mol
V = 0.750 dm3
C = 1.58 / 0.750 = 2.11 mol dm-3
Answer = 2.11 mol dm-3
n = 250 / 158.0 = 1.58 mol
V = 0.750 dm3
C = 1.58 / 0.750 = 2.11 mol dm-3
Answer = 2.11 mol dm-3
Determining the Mass
1. What mass of cobalt (II) chloride was required to be dissolved in 80cm3 of water to produce a solution that has a concentration of 0.4moldm-3?
Working Out
n = C × V = 0.4 × 0.080 = 0.032 mol
Mr(CoCl2) = 58.9 + 71.0 = 129.9
Mass = 0.032 × 129.9 = 4.16 g
(0.4 implies 1 s.f., but 80 is 2. We will use 2)
Answer = 4.2 g
Mr(CoCl2) = 58.9 + 71.0 = 129.9
Mass = 0.032 × 129.9 = 4.16 g
(0.4 implies 1 s.f., but 80 is 2. We will use 2)
Answer = 4.2 g
2. What mass of manganese (II) nitrate was required to be dissolved in 250cm3 of water to produce a solution that has a concentration of 2.5moldm-3?
Working Out
n = 2.5 × 0.250 = 0.625 mol
Mr(Mn(NO3)2) = 54.9 + 2(14.0 + 48.0) = 178.9
Mass = 0.625 × 178.9 = 111.8 g
(2.5 has 2 s.f.)
Answer = 110 g
Mr(Mn(NO3)2) = 54.9 + 2(14.0 + 48.0) = 178.9
Mass = 0.625 × 178.9 = 111.8 g
(2.5 has 2 s.f.)
Answer = 110 g
3. What mass of vanadium (III) phosphate monohydrate was required to be dissolved in 120cm3 of water to produce a solution that has a concentration of 1.5moldm-3?
Working Out
n = 1.5 × 0.120 = 0.18 mol
Mr(VPO4·H2O) = 50.9 + 31.0 + 64.0 + 18.0 = 163.9
Mass = 0.18 × 163.9 = 29.5 g
(1.5 has 2 s.f.)
Answer = 30 g
Mr(VPO4·H2O) = 50.9 + 31.0 + 64.0 + 18.0 = 163.9
Mass = 0.18 × 163.9 = 29.5 g
(1.5 has 2 s.f.)
Answer = 30 g
4. What mass of iodine was required to be dissolved in 100cm3 of water to produce a solution that has a concentration of 0.5moldm-3?
Working Out
n = 0.5 × 0.100 = 0.05 mol
Mr(I2) = 2 × 126.9 = 253.8
Mass = 0.05 × 253.8 = 12.7 g
(0.5 has 1 s.f. – or 2 if 0.50 implied, but text says 0.5)
Answer = 10 g
Mr(I2) = 2 × 126.9 = 253.8
Mass = 0.05 × 253.8 = 12.7 g
(0.5 has 1 s.f. – or 2 if 0.50 implied, but text says 0.5)
Answer = 10 g
5. What mass of chromium (III) sulfate octahydrate was required to be dissolved in 500cm3 of water to produce a solution that has a concentration of 2moldm-3?
Working Out
n = 2 × 0.500 = 1.0 mol
Mr(Cr2(SO4)3·8H2O) = 104.0 + 288.3 + 144.0 = 536.3
Mass = 1.0 × 536.3 = 536.3 g
(2 has 1 s.f.)
Answer = 500 g
Mr(Cr2(SO4)3·8H2O) = 104.0 + 288.3 + 144.0 = 536.3
Mass = 1.0 × 536.3 = 536.3 g
(2 has 1 s.f.)
Answer = 500 g
Determining the Volume
1. What volume of water was 6.0g of sodium carbonate dissolved in in order to get a concentration of 0.40moldm-3?
Working Out
Mr(Na2CO3) = 106.0
n = 6.0 / 106.0 = 0.0566 mol
V = n / C = 0.0566 / 0.40 = 0.142 dm3
(6.0 and 0.40 have 2 s.f.)
Answer = 140 cm3
n = 6.0 / 106.0 = 0.0566 mol
V = n / C = 0.0566 / 0.40 = 0.142 dm3
(6.0 and 0.40 have 2 s.f.)
Answer = 140 cm3
2. What volume of water was 400g of anhydrous potassium dichromate dissolved in in order to get a concentration of 2.0moldm-3?
Working Out
Mr(K2Cr2O7) = 78.2 + 104.0 + 112.0 = 294.2
n = 400 / 294.2 = 1.36 mol
V = 1.36 / 2.0 = 0.680 dm3
(2.0 has 2 s.f.)
Answer = 0.68 dm3
n = 400 / 294.2 = 1.36 mol
V = 1.36 / 2.0 = 0.680 dm3
(2.0 has 2 s.f.)
Answer = 0.68 dm3
3. What volume of water was 1.00kg of barium hydroxide monohydrate dissolved in in order to get a concentration of 1.60moldm-3?
Working Out
Mr(Ba(OH)2·H2O) = 137.3 + 34.0 + 18.0 = 189.3
n = 1000 / 189.3 = 5.28 mol
V = 5.28 / 1.60 = 3.30 dm3
Answer = 3.30 dm3
n = 1000 / 189.3 = 5.28 mol
V = 5.28 / 1.60 = 3.30 dm3
Answer = 3.30 dm3
4. What volume of water was 60g of sodium iodide dissolved in in order to get a concentration of 3.0moldm-3?
Working Out
Mr(NaI) = 23.0 + 126.9 = 149.9
n = 60 / 149.9 = 0.400 mol
V = 0.400 / 3.0 = 0.133 dm3
(60 has 2 s.f.)
Answer = 130 cm3
n = 60 / 149.9 = 0.400 mol
V = 0.400 / 3.0 = 0.133 dm3
(60 has 2 s.f.)
Answer = 130 cm3
5. What volume of water was 16.0g of aluminium potassium nitrate dissolved in in order to get a concentration of 0.5moldm-3?
Working Out
(Assuming AlK(NO3)4)
Mr = 27.0 + 39.1 + 4(62.0) = 314.1
n = 16.0 / 314.1 = 0.0509 mol
V = 0.0509 / 0.5 = 0.102 dm3
(0.5 has 1 s.f.)
Answer = 100 cm3 (0.1 dm3)
Mr = 27.0 + 39.1 + 4(62.0) = 314.1
n = 16.0 / 314.1 = 0.0509 mol
V = 0.0509 / 0.5 = 0.102 dm3
(0.5 has 1 s.f.)
Answer = 100 cm3 (0.1 dm3)
Dilution Factors
1. What volume of water would you need to add to a 250cm3 solution of 1moldm-3 hydrochloric acid so that its new concentration is 0.8moldm-3?
Working Out
C1V1 = C2V2
1 × 250 = 0.8 × V2
V2 = 250 / 0.8 = 312.5 cm3
Added Water = 312.5 – 250 = 62.5 cm3
(1 and 0.8 have 1 s.f.)
Answer = 60 cm3
1 × 250 = 0.8 × V2
V2 = 250 / 0.8 = 312.5 cm3
Added Water = 312.5 – 250 = 62.5 cm3
(1 and 0.8 have 1 s.f.)
Answer = 60 cm3
2. 250cm3 of 0.5moldm-3 sodium hydroxide has 100cm3 of water added to it. What is its new concentration?
Working Out
Total Volume = 350 cm3
n = 0.5 × 0.250 = 0.125 mol
C = 0.125 / 0.350 = 0.357 mol dm-3
(0.5 has 1 s.f.)
Answer = 0.4 mol dm-3
n = 0.5 × 0.250 = 0.125 mol
C = 0.125 / 0.350 = 0.357 mol dm-3
(0.5 has 1 s.f.)
Answer = 0.4 mol dm-3
3. 14.0g of sodium carbonate is dissolved in 50cm3. What volume of water do you need to add to it in order to create a solution with a concentration of 0.2moldm-3?
Working Out
n = 14.0 / 106.0 = 0.132 mol
Target V = 0.132 / 0.2 = 0.660 dm3 = 660 cm3
Added Water = 660 – 50 = 610 cm3
(0.2 has 1 s.f.)
Answer = 600 cm3
Target V = 0.132 / 0.2 = 0.660 dm3 = 660 cm3
Added Water = 660 – 50 = 610 cm3
(0.2 has 1 s.f.)
Answer = 600 cm3
4. What volume of water would you need to add to a 15cm3 solution of 2moldm-3 H2SO4 so that its new concentration is 0.6moldm-3?
Working Out
C1V1 = C2V2
2 × 15 = 0.6 × V2
V2 = 30 / 0.6 = 50 cm3
Added Water = 50 – 15 = 35 cm3
(2 and 0.6 have 1 s.f.)
Answer = 40 cm3
2 × 15 = 0.6 × V2
V2 = 30 / 0.6 = 50 cm3
Added Water = 50 – 15 = 35 cm3
(2 and 0.6 have 1 s.f.)
Answer = 40 cm3
5. What is the dilution factor if you need to decrease the concentration from 3.0moldm-3 to 0.6moldm-3?
Working Out
Dilution Factor = Initial Conc / Final Conc
DF = 3.0 / 0.6 = 5
Answer = 5
DF = 3.0 / 0.6 = 5
Answer = 5