Mole fraction
Tested in 3 real MPSC questions across 2 years (2021–2023). See it in context on the interactive Concept Graph →
Years it was asked
What you need to know first
Real questions that test this
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.
समुद्रपातळीवर असलेल्या प्रयोगशाळेत वापरण्यासाठी, वर्षाच्या कोणत्याही दिवशी खालीलपैकी सगळ्यात योग्य संहति एकक/एकके _________ हे/ही आहे/आहेत. On any day of the year, to use in laboratories at sea level the most correct concentration unit/s is/are _________.
Molality is independent of temperature because it is expressed in terms of mass of solvent (kg), whereas molarity and normality depend on volume which varies with temperature.