Problem 53
Question
\(0.5 \mathrm{M}\) of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) is diluted from 1 litre to 10 litre, normaliy of the resulting solution is (a) \(1 \mathrm{~N}\) (b) \(10 \mathrm{~N}\) (c) \(11 \mathrm{~N}\) (d) \(0.1 \mathrm{~N}\)
Step-by-Step Solution
Verified Answer
The normality of the resulting solution is \(0.1\, \mathrm{N}\).
1Step 1: Understand the Given Information
We have an initial solution of sulfuric acid \( (\mathrm{H}_2\mathrm{SO}_4) \) with a molarity of \(0.5\, \mathrm{M}\) and a volume of \(1\, \mathrm{L}\). This solution is diluted to a total volume of \(10\, \mathrm{L}\). We need to find the normality of the resulting solution.
2Step 2: Remember Relationship Between Molarity and Normality
Normality (\(N\)) for sulfuric acid is calculated as \(N = n \times M\), where \(n\) is the number of equivalents per mole of solute. For \(\mathrm{H}_2\mathrm{SO}_4\), \(n = 2\) because it can donate two protons (\(H^+\)) per molecule.
3Step 3: Calculate Initial Normality of the Acid Solution
Initially, the molarity \( (M) \) of \(\mathrm{H}_2\mathrm{SO}_4\) is \(0.5\, \mathrm{M}\). Thus, the initial normality is calculated as: \(N = 2 \times 0.5 = 1\, \mathrm{N}\).
4Step 4: Determine Normality After Dilution
Upon dilution, the number of equivalents of solute remains the same, but the concentration changes. Since normality is proportional to molarity and inversely proportional to total volume, the new normality can be calculated by: \(\text{New Normality} = \frac{\text{Initial Normality} \times \text{Initial Volume}}{\text{Final Volume}} = \frac{1 \times 1}{10} = 0.1 \mathrm{~N}\).
Key Concepts
DilutionSulfuric AcidMolarity
Dilution
When a solution is diluted, it is essentially mixed with more solvent. This process decreases the concentration of the solute in the solution, meaning that the solute is less concentrated in the larger volume. For example, if you have one liter of a 0.5 M sulfuric acid solution and you dilute it to ten liters, you are spreading the same amount of sulfuric acid over a much larger volume. This results in a decreased concentration.
You can calculate the new concentration after dilution using the formula:
You can calculate the new concentration after dilution using the formula:
- \[\text{New Normality} = \frac{\text{Initial Normality} \times \text{Initial Volume}}{\text{Final Volume}}\]
Sulfuric Acid
Sulfuric acid, with a chemical formula of \(\text{H}_2\text{SO}_4\), is a strong, diprotic acid. This means it can donate two protons or hydrogen ions (\(H^+\)) per molecule. Because of this, it behaves differently in normality calculations compared to monoprotic acids like hydrochloric acid, which can donate just one proton.
Sulfuric acid is widely used in industries and laboratories due to its strong acidic nature and ability to participate in a variety of chemical reactions. Its diprotic nature requires special consideration in calculations related to concentration:
Sulfuric acid is widely used in industries and laboratories due to its strong acidic nature and ability to participate in a variety of chemical reactions. Its diprotic nature requires special consideration in calculations related to concentration:
- Normality \((N)\) is given by: \(N = n \times M\), where \(n\) is the number of equivalents per mole of acid.
Molarity
Molarity is a measure of the concentration of a solute in a solution. It is defined as the number of moles of a solute per liter of solution and is expressed in units of mol/L, or \(\text{M}\). Molarity is crucial in chemistry because it allows chemists to know how much of a substance is present in a given volume of solution.
The formula for molarity is:
The formula for molarity is:
- \[M = \frac{\text{moles of solute}}{\text{liters of solution}}\]
Other exercises in this chapter
Problem 50
\({ }^{4} \mathbf{x}\) ' grams of water is mixed in 69 grams of ethanol Mole fraction of ethanol in the resultant solution is 0.6. what is the value of ' \(x\)
View solution Problem 51
The vapours pressure of water at \(23^{\circ} \mathrm{C}\) is \(19.8 \mathrm{~mm}\). of \(\mathrm{Hg} .0 .1\) mole of glucose is dissolved in \(178.2 \mathrm{~g
View solution Problem 55
At room temperature, the mole fraction of a solute is \(0.25\) and the vapour pressure of a solvent is \(0.80 \mathrm{~atm}\). The lowering of vapour pressure i
View solution Problem 56
\(50 \mathrm{~mL}\) of \(10 \mathrm{~N} \mathrm{H}_{2} \mathrm{SO}_{4}, 25 \mathrm{~mL}\) of \(12 \mathrm{~N} \mathrm{HCl}\) and \(40 \mathrm{~mL}\) of \(5 \mat
View solution