Titration

Tested in 3 real MPSC questions across 2 years (2021–2023). See it in context on the interactive Concept Graph →

Years it was asked

2023 2021

Real questions that test this

1.5M H₂SO₄ च्या 10 dm³ द्रावणापासून 1.0N H₂SO₄ चे द्रावण कसे बनवाल ? How is 1.0N H₂SO₄ solution is prepared from 10 dm³ of 1.5M H₂SO₄ solution ?

a)दिलेल्या द्रावणात 5 dm³ पाणी घालून / By adding 5 dm³ of water to it

b)दिलेल्या द्रावणात 294 ग्रॅम H₂SO₄ घालून / By adding 294 g of H₂SO₄ to it

c)दिलेल्या द्रावणात 20 dm³ पाणी घालून / By adding 20 dm³ of water to it&check; Correct

d)दिलेल्या द्रावणात 174 ग्रॅम H₂SO₄ घालून / By adding 174 g of H₂SO₄ to it

Normality (N) = Molarity (M) x n-factor. For H2SO4, n-factor is 2, so 1.5M H2SO4 is 3.0N. Using dilution formula N1V1 = N2V2 (3.0 x 10 = 1.0 x V2), the final volume V2 must be 30 dm³. Since we already have 10 dm³, we need to add 20 dm³ of water.

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&check; 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.

एका अनुमापनी विश्लेषणणात, 25 घनसेिंटीमीटर 0.1 N नायट्रिकआम्ल _________ घनसेिंटीमीटर 0.25 N पोटॅशिअम हायड्रॉक्साइड पूर्णपणे अभिक्रिया करेल. In a titrimetric analysis, 25 cm³ of 0.1 N HNO₃ would react completely with _________ cm³ of 0.25 N KOH.

1)100

2)10&check; Correct

3)250

4)25

Using the neutralization formula N1V1 = N2V2: (0.1 N) * (25 cm³) = (0.25 N) * V2. Solving for V2 gives (2.5 / 0.25) = 10 cm³.

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