Molarity
Molarity is an concept from the MPSC syllabus, tested in 3 past MPSC questions (2021, 2023). Below are those real questions with their correct answers and explanations.
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180 ग्रॅम अॅसेटिक आम्ल (CH3COOH) आणि 180 ग्रॅम पाणी (H2O) एकमेकांत मिसळले. तयार झालेल्या द्रावणात अॅसेटिक आम्लाचा मोल फॅक्शन ______ इतका असेल. [H = 1, C = 12, O = 16] 180 g of acetic acid (CH3COOH) is mixed with 180 g of water (H2O). The mole fraction of acetic acid in the resultant solution is [H = 1, C = 12, O = 16]
Molar mass of CH3COOH = 12+3+12+16+16+1 = 60 g/mol. Moles of CH3COOH = 180 / 60 = 3. Molar mass of H2O = 18 g/mol. Moles of H2O = 180 / 18 = 10. Mole fraction of acetic acid = Moles of acid / Total moles = 3 / (3 + 10) = 3 / 13 = 0.23.
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 = _________
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.
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.