1. Sodium citrate: 0.015 M. Calculate the mass of sodium citrate dihydrate, Na3C6H5O7(H2O)2, needed to prepare 50 mL of a 0.015 M sodium citrate solution. 2. Silver nitrate: 5.0 x 10-4 M. Calculate the volume of 0.20 M AgNO3 solution needed to prepare 50 mL of 5.0 x 10-4 M AgNO3 solution. 3. Hydrogen peroxide: 0.050 M. Calculate the volume of 1.4 M H2O2 solution needed to prepare 50 mL of 0.050 M H2O2 solution. 4. Sodium borohydride: 0.015 M. Calculate the volume of 0.15 M NaBH4 solution needed to prepare 25 mL of 0.015 M NaBH4 solution. The following four stock solutions are used for the preparation of AGNP:1. Sodium citrate: 0.015 M2. Silver nitrate: 5.0 x 104 M3. Hydrogen peroxide: 0.050 M4. Sodium borohydride: 0.015 MTo investigate the factors which affect nanoparticle formation and to optimize the formation ofnanoplates (nanoprisms), solutions 1-3 plus high purity (Millipore) water are combined inseparate test tubes according to the following table:SolutionTube aTube bTube cTube dTube eTube fTube g1O mLO mL1.0 mL1.0 mL1.0 mL1.0 mL2.0 mL22.0 mL2.0 mL2.0 mL2.0 mL2.0 mL2.0 mL2.0 mLO mL2.0 mL1.0 mL2.0 mL3.0 mL4.0 mL4.0 mLMilliporewater7.0 mL5.0 mL5.0 mL4.0 mL3.0 mL2.0 mL1.0 mLNanoparticle formation is started by the addition of 1.0 mL of solution 4 to each of the abovetubes, then briskly stirred. All reactants are colorless in solution, but once combined and after abrief induction period, a color change occurs. In some cases, the initial solution color changes toa different color or goes through multiple color changes before a final constant color is achieved.The absorbance spectrum of each solution in the range 375-950 nm is then recorded. Thesespectra are compiled and presented in part c of the activity. These experimental results wereobtained by students who performed this experiment in general chemistry lab several semestersago.

1. To calculate the mass of sodium citrate dihydrate, Na3C6H5O7(H2O)2, needed to prepare 50 mL of a 0.015 M sodium citrate solution. 2. Calculate the volume of 0.20 M AgNO3 solution needed to prepare 50 mL of 5.0 x 10-4 M AgNO3 solution. 3. Calculate the volume of 1.4 M H2O2 solution needed to prepare 50 mL of 0.050 M H2O2 solution. 4. Calculate the volume of 0.15 M NaBH4 solution needed to prepare 25 mL of 0.015 M NaBH4 solution.1. To calculate the mass of sodium citrate dihydrate, Na3C6H5O7(H2O)2, needed to prepare 50 mL of a 0.015 M sodium citrate solution. Given: Volume of solution= 50 mL Concentration of sodium citrate dihydrate, Na3C6H5O7(H2O)2 = 0.015 M Molar mass of sodium citrate dihydrate, Na3C6H5O7(H2O)2 3×MNa+6×MC+9×MH+9×MO3×22.98+6×12.01+9×1+9×15.99294.1 g/mol Moles = Concentration×Volume = 0.015 M ×50×10-3 L = 0.00075 molMass = Moles ×Molar mass = 0.00075 mol ×294.1 g/mol =0.221 g In order to prepare 50 mL of 0.015 M Sodium citrate solution, mass of sodium citrate dihydrate required is 0.221 g2. Calculate the volume of 0.20 M AgNO3 solution needed to prepare 50 mL of 5.0 x 10-4 M AgNO3 solution. Stock solution Concentration of stock solution of AgNO3 (M1) = 0.20 M Volume of stock solution AgNO3 (V1) = ? Diluted solution Concentration of Diluted solution of AgNO3 (M2)=5.0 x 10-4 M Volume of diluted solution (V2) = 50 mL Using Dilution Equation: M1V1Stock =M2V2diluted0.20 M ×V1=5.0×10-4 M×50 mLv1=5.0×10-4 M×50 mL0.20 M=0.125 mL The volume of 0.20 M AgNO3 solution required is 0.125 mL 3. Calculate the volume of 1.4 M H2O2 solution needed to prepare 50 mL of 0.050 M H2O2 solution. Stock solution Concentration of stock solution of H2O2 (M1) = 1.4 M Volume of stock solution H2O2 (V1) = ? Diluted solution Concentration of Diluted solution of H2O2 (M2)=0.050 M Volume of diluted solution H2O2 (V2) = 50 mL Using Dilution Equation: M1V1Stock =M2V2diluted1.4 M ×V1=0.050 M×50 mLv1=0.050 M×50 mL1.4 M=1.79 mL The volume of 1.4 M H2O2 solution required is 1.79 mL 4. Calculate the volume of 0.15 M NaBH4 solution needed to prepare 25 mL of 0.015 M NaBH4 solution. Stock solution Concentration of stock solution of NaBH4 (M1) = 0.15 M Volume of stock solution NaBH4 (V1) = ? Diluted solution Concentration of Diluted solution of NaBH4 (M2)=0.015 M Volume of diluted solution NaBH4(V2) = 25 mL Using Dilution Equation: M1V1Stock =M2V2diluted0.15 M ×V1=0.015 M×25 mLv1=0.015 M×25 mL0.15 M=2.5 mL The volume of 1.4 M NaBH4 solution required is 2.5 mL