Strength: Definition, Relation Between the Strength and Other Concentration Terms, Practice Problems and FAQs

Do you know what is the current strength of the Indian Parliament?

Well, the Indian parliament constitutes two houses: Lok Sabha and Rajya Sabha. The strength of Lok Sabha and Rajya Sabha is 543 and 245 respectively. Hence, the strength of the Indian Parliament will be the sum of both houses and that is 788. In this case, strength refers to the number of occupancies available in the parliament.

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Similarly, in chemistry, the strength of any solution refers to the concentration of the solution in grams per litre. Let’s understand more about strength with example.

Table of Contents

Strength of a Solution
Relation between the Strength and Other Concentration Terms
Practice Problems
Frequently Asked Questions - FAQs

Strength of a Solution

It is a measure of the concentration of the solution in terms of grams of solutes present in 1000 mL of the solution.

$S t r e n g t h o f t h e S o l u t i o n = \frac{m a s s o f s o l u t e (g)}{V o l u m e o f s o l u t i o n (L)} = \frac{m a s s o f s o l u t e (g)}{V o l u m e o f s o l u t i o n (m L)} \times 1000$

For instance, if there are 10 g of glucose (solute) dissolved in 1 L of the solution, the solution's strength will be 10 g/L.

Relation between the Strength and Other Concentration Terms

Strength of solution & molarity of solution:

$S t r e n g t h o f s o l u t i o n (S) = M o l a r i t y (M) \times M o l a r m a s s o f t h e s o l u t e$

Strength of solution & normality of solution:

$S t r e n g t h o f s o l u t i o n (S) = N o r m a l i t y (N) \times E q u i v a l e n t w e i g h t o f t h e s o l u t e$

Practice problems

Q1. Calculate the strength of the solution of NaOH, prepared by dissolving its 80g in sufficient water to form 1750 mL of the solution.

a. 0.45g L^-1
b. 45 g L^-1
c. 0.045 g L^-1
d. None of these

Answer: (C)

$S t r e n g t h o f t h e s o l u t i o n = \frac{m a s s o f s o l u t e (g)}{V o l u m e o f s o l u t i o n (L)} = \frac{m a s s o f s o l u t e (g)}{V o l u m e o f s o l u t i o n (m L)} \times 1000$

$S t r e n g t h o f t h e s o l u t i o n = \frac{80 g}{1750 L}$

=0.045 g L^-1

Q2. Which of the following is the correct unit of strength?

mol/L
g/L
mol/L-1
g/L-1

Answer: B

Strength of the solution is a measure of the concentration of the solution in terms of grams of solutes present in 1000 mL/1L of the solution. It is generally represented in g/L unit.

Q3. Calculate the strength of NaCl solution for 0.5 M NaCl solution.

a. 29.25 g L^-1
b. 58.5 g L^-1
c. 14.625 g L^-1
d. 87.75 g L^-1

Answer: (A)

Solution: Molarity = 0.5 M (given)

Molar mass of NaCl =58.5 g mol^-1

We know, $S t r e n g t h o f t h e s o l u t i o n (S) = M o l a r i t y (M) \times M o l a r m a s s o f t h e s o l u t e$

$= 0.5 M \times 58.5 g m o l^{- 1} = 29.25 g L^{- 1}$

Q 4. 150 mL of 2.5 N HCl and 300 mL of 5 N HCl solution are mixed. What will be the strength of resulting solution?

a. 84.56 g L^-1
b. 65.5 g L^-1
c. 94.76 g L^-1
d. 152.083 g L^-1

Answer: (D)

Solution: V₁= 150 mL & N₁= 2.5 N

V₂= 300 mL & N₂= 5 N

Final Normality, $N_{f} = \frac{V_{1} N_{1} + V_{2} N_{2}}{V_{1} + V_{2}}$

$= \frac{(150 \times 2.5) + (300 \times 5)}{150 + 300} = \frac{1875}{450} = 4.16 N$

Molar mass of HCl= 36.5 g mol^-1 and n-factor of HCl=1

Equivalent weight of $H C l = \frac{M o l a r m a s s o f H C l}{n - f a c t o r o f H C l} = \frac{36.5}{1} = 36.5 g e q^{- 1}$

Strength of the solution (S) = Normality (N) Equivalent mass of the solute

$= 4.16 N \times 36.5 g e q^{- 1} ≅ 152.083 g L^{- 1}$

Frequently Asked Questions - FAQs

Q1. Does the strength of the solution change with temperature?
Answer: Yes, the temperature affects the solution's strength.,

$S t r e n g t h o f t h e s o l u t i o n = \frac{m a s s o f s o l u t e (g)}{V o l u m e o f s o l u t i o n (L)}$

When temperature increases in a solution, the molecules gains kinetic energy and start moving on the random direction. This increases the volume of the solution. As the strength of the solution is inversely proportional to the volume of the solution, hence, with an increase in temperature, the strength of the solution decreases.

Q2. Is it possible to derive molality straight from the strength of the solution?
Answer: No, the strength of the solution depends on the volume of the solution.

$S t r e n g t h o f t h e s o l u t i o n = \frac{m a s s o f s o l u t e (g)}{V o l u m e o f s o l u t i o n (L)}$

Molality of the solution is calculated as

$M o l a l i t y (m) = \frac{N u m b e r o f m o l e s o f t h e s o l u t e (m o l)}{M a s s o f t h e s o l v e n t (k g)}$

Since to derive the molality of a solution we need a mass of the given solvent. The strength of the solution can provide the data for volume. To calculate mass from volume we need the value of density. Hence, it is not possible to calculate molality directly from the strength of the solution.

Q3. Is there any difference between a solution having $75 % \frac{w}{V}$ and a solution with 75 g L^-1 strength?
Answer: $75 % \frac{w}{V}$ solution, means 75 g solute present in 100 ml of the solution but 75 g L^-1 strength of solution, means 75 g solute dissolved in 1000 ml solution.

$S t r e n g t h o f s o l u t i o n = \frac{m a s s o f s o l u t e (g)}{V o l u m e o f s o l u t i o n (L)} = \frac{m a s s o f s o l u t e (g)}{V o l u m e o f s o l u t i o n (m L)} \times 1000$