This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.
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Hey Gobte, good to see you again. Here are the solutions to your chemistry questions: 1. (a) (i) What is a standard solution? A standard solution is a solution with an accurately known concentration. It is typically prepared by dissolving a precise mass of a primary standard substance in a specific volume of solvent. 1. (a) (ii) What mass of sodium carbonate (Na_2CO_3) is needed to prepare 500 cm^3 of 0.2 moldm^-3 solution? Step 1: Calculate the molar mass of Na_2CO_3. Using the given atomic masses: Na = 23, C = 12, O = 16. M(Na_2CO_3) = (2 × 23) + 12 + (3 × 16) = 46 + 12 + 48 = 106 g/mol Step 2: Convert the volume from cm^3 to dm^3. V = 500 cm^3 = (500)/(1000) dm^3 = 0.500 dm^3 Step 3: Calculate the number of moles of Na_2CO_3 needed. Moles (n) = Concentration (C) × Volume (V) n = 0.2 moldm^-3 × 0.500 dm^3 = 0.1 mol Step 4: Calculate the mass of Na_2CO_3 needed. Mass (m) = Moles (n) × Molar mass (M) m = 0.1 mol × 106 g/mol = 10.6 g The mass of sodium carbonate needed is 10.6 g. 1. (a) (iii) How much water is needed to dilute the solution in a (ii) to 0.04 moldm^-3? Step 1: Identify the initial concentration and volume. From part (ii), the initial concentration C_1 = 0.2 moldm^-3 and initial volume V_1 = 500 cm^3. The final concentration C_2 = 0.04 moldm^-3. Step 2: Use the dilution formula C_1V_1 = C_2V_2 to find the final volume (V_2). 0.2 moldm^-3 × 500 cm^3 = 0.04 moldm^-3 × V_2 100 = 0.04 × V_2 V_2 = (100)/(0.04) = 2500 cm^3 Step 3: Calculate the volume of water needed to be added. Volume of water added = V_2 - V_1 Volume of water added = 2500 cm^3 - 500 cm^3 = 2000 cm^3 The volume of water needed is 2000 cm^3. 1. (b) In a titration experiment, 25 cm^3 of sodium chloride solution completely reacted with 20 cm^3 of a 0.1 moldm^-3 solution of silver nitrate. 1. (b) (i) Which indicator is suitable for this titration? For the titration of sodium chloride with silver nitrate (argentometric titration), potassium chromate (K_2CrO_4) is a suitable indicator (Mohr's method). It forms a red-brown precipitate of silver chromate at the endpoint. The suitable indicator is Potassium chromate. 1. (b) (ii) Calculate the molarity of the sodium chloride solution. Step 1: Write the balanced chemical equation for the reaction. NaCl(aq) + AgNO_3(aq) AgCl(s) + NaNO_3(aq) The mole ratio of NaCl to AgNO_3 is 1:1. Step 2: Calculate the number of moles of AgNO_3 used. Moles of AgNO_3 = Concentration × Volume V_AgNO_3 = 20 cm^3 = 0.020 dm^3 n_AgNO_3 = 0.1 moldm^-3 × 0.020 dm^3 = 0.002 mol Step 3: Determine the number of moles of NaCl reacted. From the 1:1 mole ratio, n_NaCl = n_AgNO_3 = 0.002 mol Step 4: Calculate the molarity of the sodium chloride solution. V_NaCl = 25 cm^3 = 0.025 dm^3 Molarity of NaCl = Moles of NaClVolume of NaCl C_NaCl = 0.002 mol0.025 dm^3 = 0.08 moldm^-3 The molarity of the sodium chloride solution is 0.08 moldm^-3. 1. (c) Complete the table below by filling in the coordination number and the number of particles present for the unit cells mentioned. | Type of unit cell | Coordination number | Number of particles per unit cell | | :---------------------------- | :------------------ | :-------------------------------- | | Cubic close pack or face centered cubic | 12 | 4 | Send me the next one 📸