Sample Problems on Concentration of Solution
Problem 1: 15 g of common salt is dissolved in 400 g of water. Calculate the concentration of the solution by expressing it in Mass by Mass percentage (w/w%).
Solution:
Given that,
Mass of solute (common salt) = 15 g …(1)
Mass of Solvent (water) = 400 g …(2)
It is known that,
Mass of Solution = Mass of Solute + Mass of Solvent …(3)
So,
Substituting (1) and (2) in (3), we obtain the following,
Mass of Solution = 15 g + 400 g = 415 g …(4)
From Figure 4, we know
Mass by Mass Percentage = ( Mass of Solute / Mass of Solution ) × 100 …(5)
Substituting (1) and (4) in (5), we obtain the following,
Mass by Mass Percentage = ( 15 g / 415 g ) × 100 = ( 0.0361 ) × 100 = 3.61
Answer is:
( w / w % ) = 3.61
Problem 2: 15 g of common salt is dissolved in a solution of 300 mL, calculate the Mass by Volume percentage (w/v%).
Solution:
Given that,
Mass of solute (common salt) = 15 g . . . (1)
Mass of Solution (salt solution) = 300 mL . . . (2)
From Figure 5, we know
Mass by Volume Percentage = ( Mass of Solute / Volume of Solution ) × 100 . . . (3)
Substituting (1) and (2) in (3), we obtain the following,
Mass by Volume Percentage = ( 15 g / 300 mL ) × 100 = ( 0.05 ) × 100 = 5 g/mL
Answer is:
( w / v % ) = 5 g/mL
Problem 3: Richard dissolved 70 g of sugar in 750 mL of sugar solution. Calculate the Mass by Volume percentage (w/v%).
Solution:
Given that,
Mass of solute (common salt) = 70 g . . . (1)
Mass of Solution (salt solution) = 750 mL . . . (2)
From Figure 5, we know
Mass by Volume Percentage = ( Mass of Solute / Volume of Solution ) × 100 . . . (3)
Substituting (1) and (2) in (3), we obtain the following,
Mass by Volume Percentage = ( 70 g / 750 mL ) * 100 = ( 0.933 ) × 100 = 93.3 g/mL
Answer is:
( w / v % ) = 93.3 g/mL
Problem 4: What is the molarity of a solution containing 0.5 moles of NaCl dissolved in 500 mL of water?
Solution:
Given, Moles of NaCl = 0.5, Volume of Solution = 500 mL = 0.5 Liter
As, Molarity (M) = moles of solute / liters of solution
⇒ M = 0.5 moles / 0.5 liters = 1 M
So the molarity of the solution is 1 M.
Problem 5: What is the molality of a solution containing 20 g of glucose dissolved in 500 g of water?
Solution:
Given: Mass of Glucose = 20 g, Mass of water = 500 g = 0.5 kg,
Molar mass of glucose (C6H12O6) = 180 g/mol
Number of Moles = Mass/Molar Mass
⇒ Moles of Glucose = 20 / 180 = 1/9 ≈ 0.111 moles of glucose
As, Molality (m) = moles of solute / kilograms of solvent
⇒ m = 0.111 / 0.5 = 0.222 mol/kg
So the molality of the solution is 0.222 mol/kg.
Problem 6:How many moles of HCl are present in 250 mL of a 0.2 M HCl solution?
Solution:
Given: Molarity of solution = 0.2 M, Volume of solution = 250 mL = 0.25 liters
Molarity (M) = moles of solute / liters of solution
⇒ Moles of solute = Molarity x liters of solution
⇒ Moles of HCl = 0.2 M x 0.25 L = 0.05 moles
So there are 0.05 moles of HCl present in 250 mL of the solution.
Problem 7:What is the ppm of lead in a sample that contains 20 mg of lead in 10 L of water?
Solution:
Given : mass of solute(in mg) = 20 mg and Volution of solvent = 10 L
Mass of solution = Mass of water = 10 L x 1 Kg/L = 10 Kg (density of water is 1Kg/L or 1g/mL)
ppm (parts per million) = Mass of solute(in mg)/Mass of solution (in Kg)
⇒ ppm = 20 / 10 = 2
So the ppm of lead in the sample is 2ppm.
Problem 8: What is the ppb of mercury in a sample that contains 0.01 g of mercury in 1000 L of air?
Solution:
Given : mass of solute(in mg) = 0.01 g = 10,000 μg and Volution of solvent = 1000 L
Mass of solution = Mass of water = 1000 L x 1 Kg/L = 1000 Kg (density of water is 1Kg/L or 1g/mL)
ppb (parts per million) = Mass of solute(in μg)/Mass of solution (in Kg)
⇒ ppm = 10000 / 1000 = 10
So the ppm of lead in the sample is 10 ppb.
Concentration of a Solution
Concentration of Solution is a measure of the amount of solute that has been dissolved in the given amount of solvent. In simple words, it means determining how much of one substance is mixed with another substance. As Concentration is a frequently used term in chemistry and other relevant fields, although it is most commonly used in the context of solutions, where it refers to the quantity of solute dissolved in a solvent. Concentration can be expressed in both qualitative or quantitative (numerically) terms.
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