MPSC Rajyaseva 2021 CSAT — Question 33
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जेव्हा 4.9 ग्रॅम H₂SO₄ हे V घनसेिंटीमीटर द्रावणत असते, तेव्हा त्या द्रावणची प्रसामान्यता 0.025 N इतकी असते, तर V = _________ 4.9 g of H₂SO₄ is present in V cm³ 0.025 N solution of sulphuric acid, then V = _________
1)4
2)2000
3)4000✓ Correct
4)2
Normality = (Weight / Equivalent weight) × (1000 / V in mL). For H2SO4, equivalent weight = 49. Given Normality = 0.025, W = 4.9 g, solving gives V = 4000 cm³.