Calculations - biochemistry

Question 1

Definition of concentration

Answer 1

the amount of solute dissloved in a given amount of solvent

Question 2

Percent concentration

Answer 2

represents parts of the solute in 100 parts of total solution

amount of substance (g) dissolved in 100 g of solution

is represent in % and is independent of the MW of the substance

C(%) = (CM x MW)/10​

C(%) = (CN X Eq)/10​

​

C% = percent concentration​

CM = molar concentration​

CN = normal concentration​

Eq = equivalent weight​

MW = molecular weight

Question 3

Molarity

Answer 3

represents the number of moles of a substance per L (mol/L) or solution of millimoles per millilitre (mmol/mL)

Question 4

Normal concentration

Answer 4

represents the number of equivalent weights of a substance (Eq) per L of solution (Eq/L) or milliequivalents per millilitre (mEq/mL)

for acids: Eq = MW of the acid/number of protons exchanged in a neutralization reaction​

for bases: Eq = MW of the base/number of OH or NH2 groups exchanged in a neutralization reaction​

for salts: Eq = MW of a salt/(number of metal atoms x valence of the metal)​

Question 5

How much 72% (v/v) ethyl alcohol is needed to prepare 1200 mL of 35% (v/v) solution?

Answer 5

72/100 x X = 1200 mL x 35/100

X = 583,33 mL

or

1) 1200 mL of 35% (35% → 35 mL ….. 100 mL)

35 mL ….. 100 mL

X mL ….. 1200 mL

X = 420 mL

2) How much 72% (v/v) ethyl alcohol is needed:

72 mL ….. 100 mL

420 mL ….. Y

Y = 583.3 mL

Question 6

How much water should be added to 800 mL of 85% (v/v) certain solution to make a 40% (v/v) solution?

Answer 6

40/100 x X = 800 mL x 85/100

X = 8 x 85 x 100/40 = 1700 mL

or

1) 800 mL of 85% (85% → 85 grams ….. 100 mL

85 mL ….. 100 mL

X mL ….. 800 mL

X = 680 mL

2) … to make a 40% (v/v) solution

40 mL ….. 100 mL

680 mL….. Y mL

Y = 1700 mL

Question 7

A solution is 16% (w/w) NaOH. How many grams of NaOH are in 200 g of a solution?

Answer 7

m (NaOH) = w x m (solution)

= 16/100 x 200 g = 32 g

or

16% NaOH → 16 grams ….. 100 grams

16 grams ….. 100 grams

X grams ….. 200 grams

X = 32 grams

Question 8

Calculate the percent concentration in w/v when dissolving 120 g NaCl into a certain amount of water to make 3 L of solution.

Answer 8

C% = msolute/ msolution x 100

or

X grams ….. 100 mL

120 grams ….. 3000 mL

w/v percent = 4%

C% (w/v) = (120 g/3000 mL) x 100% = 4%

Question 9

How many mL of water are needed to produce a 5.5% solution (w/v) with 26 g of salt?

Answer 9

m (solution) = m (salt) / w

m (solution) = 26 g / 5,5/100 = 472,73 mL

or

5. 5% solution (w/v) → 5.5 grams ….. 100 mL
6. 5 grams ….. 100 mL

26 grams ….. X

X = 472.73 mL

Question 10

A solution has a volume of 375 mL and containes 42.5 g of NaCl. What is its molarity?

Answer 10

C_M = m_solute / M x m_solution

MW_NaCL = 58,44 g/mol

1) C% = msolute/ msolution x 100 = 42,5 g / 375 mL x 100 = 11,3 %
2) C_M(NaCl) = C% x 10/ MW = 11,3 % x 10 / 58,44g/mol = 1,94

or

42.5 g NaCl ….. 375 mL

X g NaCl ….. 100 mL

X = 11.33g NaCl

MW NaCl = 58.44

C% = (CM x MW)/10

CM = (C% x 10)/MW = (11.33 x 10)/58.44 = 1.94

Question 11

You dissolve 152.5 g of CuCl2 in water to make a solution with a final volume of 2.25 L. What is its molarity?

Answer 11

MW(CuCl₂) = 134,45 g/mol

1) C% = m_solute/ m_solution x 100

= (152,5 g / 2250 mL) x 100 = 6,78 %

2) C_M = (C% x 10) / MW = (6,78 % x 10) / 134,45 g/mol x 100 = 0,504

or

152.5 grams CuCl2 ….. 2.25 L (2250 mL)

X grams CuCl2 ….. 100 mL

X = 6.78

C% = (CM x MW)/10

CM = (C% x 10)/MW = (6.78 x 10)/134.45 = 0.504

Question 12

Calculate the molarity of a solution prepared by dissolving 9.8 moles of solid NaOH in enough water to make 3.62 L of solution.

Answer 12

1) m = n x M; MW(NaOH) = 39,997 g/mol

m = 9,8 mol x 39,997 g/ mol = 391, 97 g

2) C% = (m_solute / m_solution) x 100 = (391,97 g / 3620 ml) x 100 = 10,83 %
3) C_M(NaOH) = C% X 10 / MW = (10,83 x 10) / 39,997 g/mol

= 2,71

or

9.8 moles of solid NaOH ….. 3620 mL

X moles of solid NaOH ….. 1000 mL

X = 2.71 moles → CM = 2.71

Question 13

Calculate the molar/mM and % concentration of a 3N Ca(OH)2 solution.

Answer 13

MW(Ca(OH)₂ = 74,093 g/mol

1) Eg(Ca(OH)₂) = MW/n = 74,093 g/mol/ 2 = 37,0465 g/mol
2) C% = (CN x Eq) / 10 = (37,0465 g/mol x CN)/ 10 = (37,0465 g/mol x 10) / 2 = 11,12g/mol
3) CM = (C% x 10) / MW = (11,12 g/mol x 10) / 74,093 g/mol = 1,5 M = 1500 mM

Question 14

Calculate the amount in grams necessary to prepare 325 mL of a:

15% NaOH solution, 100 mM Ca(OH)2, 5N Ca(OH)2.

Answer 14

MW(Ca(OH₂) = 74,09 g/mol

1) Eg = M/n = 74,09 g/mol / 2 = 37,045 g/mol
2) C% = (CN x Eg) / 10 = (5 x 37,045) / 10 = 18,52%
3) m_solute = C% x m_solution / 100 = (18,52% x 325 ml) / 100 = 60, 19 ml

Question 15

Calculate the amount in grams of Mg(OH)2 which is necessary to prepare 330 mL of a 15% Mg(OH)2 solution. (MW Mg(OH)2 = 58)

Answer 15

1) C_M = (C% x 10)/ MW = (15 x 10) / 58 g/mol = 2,59 mol/g
2) m_solute = MW x V_solution x C_M = 58 g/mol x 0,330 L x 2,59 mol/g = 49, 5726 L ?
* C_M = (C% x 10)/ MW = m_solute / (M x Vsolution)*