# Chemistry Questions and Answers – Expressing Concentration of Solutions

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This set of Chemistry Multiple Choice Questions for Schools focuses on “Expressing Concentration of Solutions”.

1. What is the normality of lead (II) nitrate if the density of its 26% (w/w) aqueous solution is 3.105 g/mL? Take molar mass of lead (II) nitrate to be 331g/mol.
a) 2.437 N
b) 4.878 N
c) 0.243 N
d) 0.488 N

Explanation: Consider 100g of solution. It is made up of 26g lead (II) nitrate and 74g water.
Volume of solution, V = 100g/(3.105g/ml) = 32.2061 ml = 0.0322 L
Equivalent weight of lead nitrate = 331/2 = 165.5 g/eq
Number of equivalents, N = 26g/(165.5g/eq) = 0.1571 eq
Normality = N/V = 4.878 N.

2. What is the mole fraction of glycerin C3H5(OH)3 in a solution containing 33g of glycerin, 60g isopropyl alcohol and rest water?
a) 0.359
b) 0.258
c) 0.205
d) 0.480

Explanation: Mw of glycerin = 92g/mol. Number of moles of glycerin = 33/92 = 0.3587 mol
Number of moles of isopropyl alcohol = 60g/(60g/mole) = 1 mole
Number of moles of water = 7g/(18g/mole) = 0.3889 mole
Mole fraction of glycerin = 0.3587/(0.3587 + 1 + 0.3889) = 0.205.

3. If a urea (NH2CONH2) contains 45% (by mass) N2, what is the actual urea content in the sample?
a) 103.7 kg
b) 96.4 kg
c) 9.65 kg
d) 10.4 kg

Explanation: Let us consider 100 kg sample or urea. This is said to contain 45kg N2.
Now, 60kg urea contains 28 kg N2. (Since, Mw of urea = 14 + 2 + 12 + 16 + 14 + 2 = 60kg/kmole)
Then, 1kg urea contains $$\frac{28}{60}$$kg N2.
Using unitary method, x kg urea will contain 45 kg N2.
Solving for x –
x= 45 x $$\frac{60}{28}$$ = 96.4 kg.
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4. Iron (III) oxide chunks contain 80 ppm silica (SiO2). What is the concentration of this impurity in mass%?
a) 0.008%
b) 0.080%
c) 0.800%
d) 8.000%

Explanation: 80 ppm silica means there are 80 mg of silica is 106 mg (or 1 kg) of iron (III) oxide and silica mixture. The mixture can be regarded as 106 mg (or 1 kg) of iron (III) oxide since 80<<106
Mass% = (mass of impurity/mass of total mixture) x 100%
Mass% = (80mg/106mg) x 100% = 0.008%.

5. Consider 100 ml of 0.3 molar solution formed by dissolving 3.33g of XCl2 in water. What is the molar mass of element X? (Atomic mass of Cl = 35.5)
a) 9
b) 24
c) 40
d) 87

Explanation: Given,
Concentration, c = 0.3 mole/L
Volume, v = 100 ml = 0.1 L
Number of moles, n = concentration (mole/L) x volume (L)
n = 0.3 mole/L x 0.1 L = 0.03 moles
0.03 moles of XCl2 corresponds to 3.33g, then 1 mole contains 3.33/0.03 = 111 g of XCl2.
Hence, molecular mass = 111 g/mole.
Let x be the atomic mass. Then –
111 = x + 2(35.5)
x = 40g/mole.

6. What is the molality of a dilute aqueous 0.02 N H3PO4 solution?
a) 0.0050
b) 0.0200
c) 0.00330
d) 0.0067

Explanation: Given,
Normality, N = 0.02
It is known that N = nf x M, where nf is the n-factor of equivalence and M is the molarity of the solution.
On expanding, N = nf x $$\frac{Number \, of \, moles}{volume \, of \, solution(L)}$$
For H3PO4, nf = 3 since there are 3 breakable OH bonds.
Therefore, N = 3 x $$\frac{Number \, of \, moles}{mass \, of \, solvent(kg)}$$ . Here, volume of solution (1 L) can be approximated as mass of solvent (1 kg) since it is an dilute solution in water (density = 1 kg/L) implying water concentration is far greater such that amount of H3PO4 is negligible.
Now, $$\frac{Number \, of \, moles}{mass \, of \, solvent(kg)}$$ = molality of solution
Thus, N/3 = 0.02/3 =0.0067 molal.

7. What is the molarity of a 15 ml, 2 M aqueous solution when 285 ml of water is added to it?
a) 0.400 M
b) 0.100 M
c) 0.111 M
d) 0.105 M

Explanation: Given,
Initial concentration, M1 = 2 M
Initial total volume, V1 = 15 ml
Final total volume, V2 = 15 + 285 = 300 ml
Final concentration, M2 = to be found
Before and after adding water number of moles of solute remain constant since it does not react with water.
Number of moles = concentration x volume
Thus, M1 x V1 = M2 x V2
On substituting, M2 = (2 x 15)/300 = 0.1 M.

8. Calculate the volume (mL) of concentrated acid required to prepare 500 mL of 0.25 N HCl solution from concentrated stock HCl solution (specific gravity = 1.19) and 37.2% (by mass).
a) 12.128 mL
b) 20.613 mL
c) 10.307 mL
d) 24.256 mL

Explanation: Given,
Final concentration, N2 = 0.25 N
Final volume, V2 = 500 mL = 0.5 L
Consider 100 kg of solution.
100 kg of stock solution contains 37.2 kg acid.
Volume of stock solution containing 37.2 kg acid = 100kg/(1.19 kg/L) = 84.0336 L
Number of moles of HCl acid, n = mass of acid/molar mass of acid
n = 37.2 kg/(1 + 35.5) = 1.01918 kmole = 1019.18 mole
Molarity, M = n/volume of stock solution = 1019.18/84.0336 = 12.1282 M
Normality of stock solution, N1 = nf x Molarity = 1 x 12.1282 = 12.1282 N
Using equivalence equation N1 x V1 = N2 x V2:
Volume of stock solution to be added, V1 = (0.25 x 500 mL)/(12.1282) = 10.307 mL.

9. Calculate the mole fraction of A if 25g of it is dissolved in 50 moles of B. Given the molar mass of A is 25 g/mole.
a) 0.0196
b) 0.5000
c) 0.3333
d) 0.9259

Explanation: Given,
Mass of A, mA = 25g
Molar mass of A = 25 g/mole
Number of moles of A, nA = mA/MA = 25/25 = 1 mole
Number of moles of B, nB = 50 moles
Mole fraction of A, xA = nA/(nA+nB) = 1/(1 + 50) = 0.0196.

10. Which of the following concentration determine techniques gives a more accurate value?
a) Molarity
b) Molality
c) Formality
d) Normality

Explanation: Molality gives the more accurate reading since it is a temperature independent term. On the other hand, the other terms are temperature dependent since they make use of volume of solution, which tends to change with changing temperatures.

11. In a saturated solution with endothermic dissolution, how does the concentration of dissolved solute change with increasing then decreasing temperature?
a) Keeps increasing
b) Keeps decreasing
c) Increases and decreases
d)Decreases and increases

Explanation: When a solution is saturated a thermodynamic equilibrium exists between the undissolved solute and dissolved solute. Since the dissolution process here is mentioned as endothermic increasing and decreasing temperatures will decrease and increase the concentration, respectively, as given by Le Chatelier’s Principle of equilibrium.

12. What symbol is used to denote ‘molality’?
a) M
b) m
c) mM
d) n

Explanation: Universally, ‘m’ is used to denote molality concentrations. E.g. 2 m means a solution having concentration 2 molal (2 moles of solute/kg solvent). ‘M’ is used to molarity. E.g. 2 M means a solution having concentration 2 molar (2 moles of solute/L solution). ‘mM’ corresponds to milli-molar concentration and ‘n’ denotes number of moles of a component.

13. What is the molality of a solution formed when 58.5g of NaCl is dissolved in 2000 mL of water?
a) 29.2500 m
b) 0.5000 m
c) 0.0005 m
d) 0.2925 m

Explanation: Given,
Mass of NaCl, m = 58.5 g
Moles of NaCl, n = mass/molar mass
n = 58.5g/(58.5 g/mole) = 1 mole
Molality = moles of solute/mass of solvent (in kg)
Volume of solvent = 2000 ml = 2 L
Mass of solvent = volume x density = 2 L x 1 kg/L = 2 kg
Molality = 1 mole/2 kg = 0.5 molal.

14. What is the molarity of 20% (w/v) H2SO4 solution?
a) 0.51 M
b) 1.02 M
c) 4.08 M
d) 2.04 M

Explanation: Given,
20% (w/v) means 20 g of H2SO4 is present in 100 mL (= 0.1L) of solution.
Data –
Moles of solute (H2SO4) = mass of solute/molar mass
Moles of solute, n = 20 g/(2 x 1 + 32 + 4 x 16) g/mole = 0.204 mole
Volume, V = 0.1 L
Molarity, M = moles of solute (n)/volume of solution (V)
M = 0.204 moles/0.1 L = 0.204 M.

15. What is a solution called when the concentration of the solute equals its solubility in the solvent?
a) Dilute
b) Saturated
c) Unsaturated
d) Supersaturated

Explanation: Saturated solutions are those which contain maximum amount of solute that is soluble and cannot be further dissolved. Dilute solutions have a very low concentration of the salt. Unsaturated solutions have concentration of solute lesser than its solubility. Supersaturated solutions contain more amount of solute than how much is soluble in the solvent. In other words, adding more solute to a saturated solution results in a supersaturated solution.

16. If 2 L, 4 L and 6 L of three separate solutions of concentrations 1 M, 2 M and 3 M, respectively, are mixed together then what is the concentration of the resultant mixture?
a) 2.333 M
b) 6.000 M
c) 3.333 M
d) 1.500 M

Explanation: Given,
V1 = 2 L
V2 = 4 L
V3 = 6L
M1 = 1 M
M2 = 2 M
M3 = 3 M
Resultant concentration, MR = (M1V1 + M2V2 + M3V3)/(V1 + V2 + V3)
MR = ( 2 x 1 + 4 x 2 + 6 x 3 )/( 2 + 4 + 6) = 2.333 M.

17. What is the number of moles in 650 mL of 98% (v/v) sulfuric acid solution if density of sulfuric acid is 1.83 g/cm3?
a) 63.7
b) 11.9
c) 6.5
d) 11.6

Explanation: 98% (v/v) means 98 mL of acid is present in 100 mL of solution.
In 650 mL of solution, volume of acid present = 0.98 x 650 mL = 637 mL
Mass of acid present = volume x density = 637 mL x 1.83 g/mL =1165.71 g
Molar mass of sulfuric acid (H2SO4) = (2 x 1 + 32 + 4 x 16) = 98 g/mole
Number of moles of sulfuric acid = mass/molar mass = 1165.71g/(98 g/mole) = 11.9 moles.

18. What does the unit ‘mmole/kg’ represent?
a) Molarity
b) Molality
c) Molar mass
d) Milli-molarity

Explanation: The base of calculating molality of a solution is finding out the number of moles of solute per kg of solvent. Molarity is the number of moles o solute present per unit volume of solution, in liters. Molar mass refers to the mass of substance constituting one mole. Milli-molarity refers to number of milli-moles of solute present per unit volume of solution, in liters.

19. What is the concentration, in ppm, if 0.025 g of KCl is dissolved in 100 grams of water?
a) 4 x 103 ppm
b) 250 ppm
c) 2.5 x 10-4 ppm
d) 2.5 ppm

Explanation: ppm = (mass of solute/mass of solution) x 106
Mass of solute = 0.025 g
Mass of solution = 0.025 + 100 = 100.025 g
Concentration in ppm = (0.025/100.025) x 106 = 250 ppm.

20. Which of the following may be the most appropriate unit of expressing lead concentrations in drinking water?
a) mg
b) mmole
c) ppm
d) ppb

Explanation: Drinking water should be absolutely clean from all sorts of toxic, hazardous elements. Trace quantities may cay cause poisoning in the human body. Universally, it has been considered appropriate to express lead concentrations in ppb (parts per billion) for drinking water.

Sanfoundry Global Education & Learning Series – Chemistry – Class 12.

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