BITSAT Solutions — practice questions
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Practice BITSAT Solutions in the app →Which statement correctly describes Raoult's law for an ideal solution of two volatile liquids A and B?Colligative properties of a dilute solution depend on:
The figure shows a non-ideal solution with *The molal boiling-point elevation constant of water is $K_b = 0.52$ K kg/mol. What is the boiling point of a s$36$ g of glucose ($M = 180$ g/mol) is dissolved in $200$ g of water. The molality of the solution is:The van't Hoff factor $i$ for a $0.01$ M aqueous solution of NaCl, assuming complete dissociation, is approximMolarity M is defined as:Molality m is defined as:Mole fraction x_A is:Raoult's law for an ideal solution (one component A in solvent S) is:Henry's law for gas dissolution: solubility of a gas (mass m_gas per unit mass of solvent) is:Colligative properties depend on:What is the molarity of a solution containing 5.85 g of NaCl (M.W. = 58.5) in 250 mL of solution?Boiling point elevation: ΔT_b = K_b × m. For 1 molal aqueous urea (K_b = 0.52 K kg/mol), ΔT_b equals:Freezing point depression: ΔT_f = K_f × m. For 0.5 molal NaCl (assume full dissociation, K_f = 1.86), the deprVapour pressure of pure water at 25°C is 23.8 mm Hg. The vapour pressure of a solution with 18 g glucose (M.W.Osmotic pressure π = nRT/V (or CRT for molarity C). What is π for 0.1 M sugar solution at 27°C (R = 0.0821)?The van't Hoff factor i for a strong electrolyte like CaCl2 (full dissociation) is:In ideal solutions, ΔH_mix is:Mass percentage (w/w) of a solution: 12 g sugar in 100 g of solution is:Two solutions are isotonic if:For an azeotrope of HCl and water (boiling at 110°C with 20.2% HCl by mass), the composition and boiling pointMaximum boiling azeotropes are formed by mixtures with:Molality of a solution does not depend on:For 0.5 m solutions of NaCl, KCl, and BaCl2, which gives largest ΔT_f?Pure water boils at 100°C; pure ethanol at 78°C. A 0.5 mole fraction mixture under positive deviation will:Osmotic pressure of 0.1 M Glucose at 27°C is π_g. For 0.1 M NaCl at same T, the osmotic pressure is approximatThe pKa of a weak acid HA is 4.75. Find pH of an equimolar mixture of HA and A⁻:Solubility of N2 in water at 25°C is 6.8 × 10⁻⁴ mol/L at 1 atm partial pressure. By Henry's law, at 5 atm partTwo solutions A (0.1 M urea, π_A) and B (0.1 M NaCl, π_B). Water flows by osmosis:A solution boils at 100.5°C (water at 100). If K_b = 0.52, molality is:Mole fraction of solute in 18% glucose (180 g/mol) by mass in water:Vapour pressure of pure water is 100 (arbitrary units). Add 1 mole non-volatile solute to 9 moles water. New vA solution is:Mole fraction of solute x_B in a binary solution:Molarity M is defined as:Molality m is defined as:Henry's law states partial pressure of a gas above liquid is proportional to:Raoult's law for ideal solution states vapor pressure of solvent:For dilute solutions, lowering of vapor pressure of solvent ΔP / P°_A =Elevation in boiling point ΔT_b for solution of non-volatile solute:Depression in freezing point ΔT_f for non-electrolyte solute:Osmotic pressure π for dilute solution:For NaCl in water (assumes complete dissociation), van't Hoff factor i =For 0.1 M urea in water, freezing point depression (K_f = 1.86):Mass percent of solute = (mass of solute / mass of solution) × 100. For 5 g NaCl in 95 g water:Two volatile liquids A (P°_A = 100) and B (P°_B = 200) form ideal solution. For x_A = 0.4, total vapor pressurOsmotic pressure of 0.01 M sucrose solution at 27°C (R = 0.0821 L atm/mol K):For solution of urea (M_w = 60 g/mol) with 6 g in 100 g water, molality:For 0.1 M acetic acid (Ka = 1.8 × 10⁻⁵), van't Hoff factor i is:For salt CaCl₂ (assume full dissociation), i =For a 1 m glucose aqueous solution (i = 1), boiling point:Freezing-point depression and elevation are colligative properties: depend on: