Question:
1 g of fuming H\(_2\)SO\(_4\) (oleum: mixture of H\(_2\)SO\(_4\) and SO\(_3\) having formula H\(_2\)S\(_2\)O\(_7\)) is diluted with H\(_2\)O. This solution is completely neutralised by 26.7 mL of 0.8 N NaOH. Find the percentage of free SO\(_3\) in the oleum.
1 g of fuming H\(_2\)SO\(_4\) (oleum: mixture of H\(_2\)SO\(_4\) and SO\(_3\) having formula H\(_2\)S\(_2\)O\(_7\)) is diluted with H\(_2\)O. This solution is completely neutralised by 26.7 mL of 0.8 N NaOH. Find the percentage of free SO\(_3\) in the oleum.
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Oleum is H\(_2\)SO\(_4\) + SO\(_3\). Equivalent weight of SO\(_3\) = 80/2 = 40.
Updated On: Apr 23, 2026
- 20.73%
- 43.80%
- 79.27%
- 60.74%
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Concept:
Oleum contains H\(_2\)SO\(_4\) and SO\(_3\). SO\(_3\) reacts with water to form H\(_2\)SO\(_4\). Total acidity = from both. Equivalent weight of SO\(_3\) = 40 (M/2).
Step 2: Detailed Explanation:
Let H\(_2\)SO\(_4\) = x g, SO\(_3\) = (1 - x) g.
Eq. of H\(_2\)SO\(_4\) = x/49, Eq. of SO\(_3\) = (1-x)/40.
Total Eq. = x/49 + (1-x)/40.
NaOH used = 26.7 \(\times\) 0.8 / 1000 = 0.02136 eq.
x/49 + (1-x)/40 = 0.02136
Multiply by 1960: 40x + 49(1-x) = 41.8656 \(\Rightarrow\) 40x + 49 - 49x = 41.8656 \(\Rightarrow\) -9x = -7.1344 \(\Rightarrow\) x = 0.7927 g H\(_2\)SO\(_4\).
SO\(_3\) = 1 - 0.7927 = 0.2073 g \(\Rightarrow\) 20.73%.
Step 3: Final Answer:
Thus, percentage of free SO\(_3\) = 20.73%.
Oleum contains H\(_2\)SO\(_4\) and SO\(_3\). SO\(_3\) reacts with water to form H\(_2\)SO\(_4\). Total acidity = from both. Equivalent weight of SO\(_3\) = 40 (M/2).
Step 2: Detailed Explanation:
Let H\(_2\)SO\(_4\) = x g, SO\(_3\) = (1 - x) g.
Eq. of H\(_2\)SO\(_4\) = x/49, Eq. of SO\(_3\) = (1-x)/40.
Total Eq. = x/49 + (1-x)/40.
NaOH used = 26.7 \(\times\) 0.8 / 1000 = 0.02136 eq.
x/49 + (1-x)/40 = 0.02136
Multiply by 1960: 40x + 49(1-x) = 41.8656 \(\Rightarrow\) 40x + 49 - 49x = 41.8656 \(\Rightarrow\) -9x = -7.1344 \(\Rightarrow\) x = 0.7927 g H\(_2\)SO\(_4\).
SO\(_3\) = 1 - 0.7927 = 0.2073 g \(\Rightarrow\) 20.73%.
Step 3: Final Answer:
Thus, percentage of free SO\(_3\) = 20.73%.
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