please answer me for 3 and 4 thanks Part L Preparing solutions from solidsTo load the assignment, select "Load an Assignment" from the "File" menų, and selectMolarity and Density: Making Solutions from SolidsFor the following problems, calculate the mass of solid needed to make the solution, and thenprepare the solutions in the Virtual Lab. Use the solution viewer to verify that the solutions youcreated have the target concentrations. Outline the steps you took in making each solution and inverifying that it has the correct concentration.1) 100 ml of solution that is 1.0 M in sodium cation, using sodium chloride.2) 100 ml of a solution that is 0.7 m in chloride anion, using magnesium chloride.3) 250 ml of a solution that is 0.34 M in sodium cation, using sodium sulfate decahydrate.4) 500 ml of a solution that is 0.25 m in sulfate anion, using sodium sulfate decahydrate..docx44.jpg4.jpghp

Molarity is defined as the number of moles of solute present in one litre of solution. Mathematically it is shown as: molarity=number of moles of solutevolume of solution in L.Since,number of moles of solute=given mass of solutemolar mass of soluteHence, molarity of a solution = given mass of solutemolar mass of solute×1volume of solution in L. Molality of a solution is defined as the number of moles of solute present in one kg of solvent. molality=number of moles of solutemass of solvent in kg3. Molecular formula of sodium sulfate decahydrate: Na2SO4.10H2O Volume of solution = 250mL Required concentration of sodium cation in the solution is 0.34M. Since, 1mol. of Na2SO4.10H2O has ------ 2mol of sodium cations For one litre solution, 0.34M sodium cations present in --- 0.17mol of Na2SO4.10H2O. But the volume given is 250mL. For 1000mL -------0.17molNa2SO4.10H2O then, for 250mL -------- ? mol of Na2SO4.10H2O ⇒250mL×0.17mol Na2SO4.10H2O1000mL=0.0425mol of Na2SO4.10H2ORequired number of moles of Na2SO4.10H2O is------- 0.0425mol. Mass of Na2SO4.10H2O in 0.0425mol is: mass of a substance= number of moles of the substance × its molar mass =0.0425mol×322.2g/mol = 13.69g. Hence, mass of sodium sulfate decahydrate required is -------- 13.7g.4) Molality of a solution is defined as the number of moles of solute present in one kg of solvent. molality=number of moles of solutemass of solvent in kg Molecular formula of sodium sulfate decahydrate is:Na2SO4.10H2O 1 mole of sodium sulfate decahydrate has 1 mole of sulfate ions. For 1 kg solvent, 0.25mol. of sulfate ions are present in 0.25mol of sodium sulfate decahydrate. In case of water, density is 1.00g/ml at 25oC. So, 500mL water weighs------- 500g. of mass. For 1000g of solvent requires -------- 0.25mol of sodium sulfate decahydrate. then, 500g of solvent requires ----------- how many moles of sodium sulfate decahydrate? ⇒500g.×0.25mol1000g.=0.125mol Hence, 0.125mol of sodium sulfate decahydrate is required.Mass of 0.125mol. of sodium sulfate decahydrate is: mass of a substance=number of moles ×its molar mass =0.125mol×322.2g/mol =40.275g =40.3g (to three significant figures) Hence,40.3g of sodium sulfate decahydrate is required to make 500ml of 0.25m sulfate ions.Answers: 3) 13.7 g of sodium sulfate decahydrate can make 250mL of 0.34M sodium ions solution. 4) 40.3g of sodium sulfate decahydrate can make the solution of 500ml and 0.25m sulfate ions.