What is the amount of anhydrous Na2CO3 in 200 ml of 0.25 M solution? (Options: a) 6.225 g b) 66.25 g c) 6.0 g d) 6.625 g)

Here are the solutions to the multiple-choice questions: Q.1 (i) The amount of anhydrous Na_2CO_3 present in 200 ml of 0.25 M solution is. Step 1: Calculate the molar mass of Na_2CO_3. Molar mass of Na_2CO_3 = (2 × 23.0) + 12.0 + (3 × 16.0) = 46.0 + 12.0 + 48.0 = 106.0 g/mol. Step 2: Calculate the moles of Na_2CO_3. Given Molarity (M) = 0.25 mol/L Given Volume (V) = 200 mL = 0.200 L Moles = M × V = 0.25 mol/L × 0.200 L = 0.050 mol Step 3: Calculate the mass of Na_2CO_3. Mass = Moles × Molar mass = 0.050 mol × 106.0 g/mol = 5.3 g Looking at the options, none of them exactly match 5.3 g. Let's recheck the calculation and options. (a) 6.225 g (b) 66.25 g (c) 6.0 g (d) 6.625 g There might be a slight discrepancy in the question or options provided. However, if we assume a typo and look for the closest option, 6.0 g (c) is the closest to 5.3 g. Let's re-evaluate if there's any common mistake or alternative interpretation. If the molarity was 0.25 M and volume 250 mL (0.25 L), then moles = 0.25 0.25 = 0.0625 mol. Mass = 0.0625 106 = 6.625 g. This matches option (d). It's possible the volume was intended to be 250 mL instead of 200 mL, or the molarity was different. Given the options, 6.625 g is a direct calculation if the volume was 250 mL. Let's assume the question intended for 250 mL or there's a typo in the options. If we strictly follow 200 mL, the answer is 5.3 g. Since 6.625 g is an option, it's highly likely that the volume was meant to be 250 mL. Let's assume the volume is 250 mL for option (d) to be correct. Moles = 0.25 mol/L × 0.250 L = 0.0625 mol Mass = 0.0625 mol × 106.0 g/mol = 6.625 g Given the options, this is the most plausible intended answer. The final answer is (d) 6.625 g. Q.1 (ii) Molarity of pure water is Step 1: Assume a volume of pure water, e.g., 1 L. Density of water is approximately 1 g/mL. Mass of 1 L (1000 mL) of water = 1000 mL × 1 g/mL = 1000 g. Step 2: Calculate the molar mass of water (H_2O). Molar mass of H_2O = (2 × 1.0) + 16.0 = 18.0 g/mol. Step 3: Calculate the moles of water in 1 L. Moles = MassMolar mass = 1000 g18.0 g/mol ≈ 55.55 mol Step 4: Calculate the molarity. Molarity = Moles / Volume (L) Molarity = 55.55 mol1 L = 55.55 M The final answer is (d) 55.5 M. Q.1 (iii) If 4g of NaOH are dissolved in 500 ml of water, the concentration of the solution will be. Step 1: Calculate the molar mass of NaOH. Molar mass of NaOH = 23.0 + 16.0 + 1.0 = 40.0 g/mol. Step 2: Calculate the moles of NaOH. Given mass of NaOH = 4 g. Moles = MassMolar mass = 4 g40.0 g/mol = 0.1 mol Step 3: Calculate the molarity of the solution. Given volume of solution = 500 mL = 0.500 L. Molarity = MolesVolume (L) = 0.1 mol0.500 L = 0.2 M The final answer is (b) 0.2 M. Q.1 (iv) Which of the following has the minimum vapour pressure? Vapour pressure is a colligative property, which means it depends on the number of solute particles in the solution. A higher concentration of solute particles leads to a lower vapour pressure. We need to calculate the effective concentration (i × M) for each option, where i is the van't Hoff factor and M is the molarity. (a) 0.1 M BaCl_2 solution: BaCl_2 dissociates into Ba^2+ and 2Cl^-, so i = 3. Effective concentration = 3 × 0.1 M = 0.3 M. (b) 0.1 M Phenol solution: Phenol (C_6H_5OH) is a non-electrolyte (or very weak acid, usually considered non-dissociating for colligative properties). So, i = 1. Effective concentration = 1 × 0.1 M = 0.1 M. (c) 0.1 M sucrose solution: Sucrose (C_12H_22O_11) is a non-electrolyte. So, i = 1. Effective concentration = 1 × 0.1 M = 0.1 M. (d) 0.1 M solution of sodium chloride (NaCl): NaCl dissociates into Na^+ and Cl^-, so i = 2. Effective concentration = 2 × 0.1 M = 0.2 M. Comparing the effective concentrations: (a) 0.3 M (b) 0.1 M (c) 0.1 M (d) 0.2 M The 0.1 M BaCl_2 solution has the highest effective concentration (0.3 M), and therefore it will have the minimum vapour pressure. The final answer is (a) 0.1 M BaCl_2 solution. Q.1 (v) Benzene and toluene form a Benzene and toluene are structurally similar non-polar organic compounds. When mixed, the intermolecular forces between benzene-benzene, toluene-toluene, and benzene-toluene molecules are very similar. This leads to a solution that closely obeys Raoult's law. The final answer is (a) Ideal solution. Q.1 (vi) Which of the following solutions has the highest osmotic pressure? Osmotic pressure () is a colligative property, directly proportional to the effective molar concentration (i × C), where i is the van't Hoff factor and C is the molar concentration. All options have a concentration of 1 M. We need to find the solution with the highest i value. *