3Dilution is a process in which the concentration of a solution is lowered by adding more solvent. Some of the reasons dilutions are performed are to minimize measurement errors whenpreparing a series of solutions at different concentrations, to save time and laboratory space, and to make more accurate measurements on an analytical balance when the targetconcentration is very low. The new concentration of a diluted solution can be determined from the following equation, sometimes called the dilution equation,YM₁ V₁ = M₂V₂where M₁ is the concentration of the initial solution, V₁ is the volume of the initial solution, M₂ is the concentration of the final solution, and V₂ is the volume of the final solution. Theterms M₁ and M₂2 are often expressed in molarity, M, which is equal to mol/L. Another way to write the equation is to replace subscripts of 1 and 2 with abbreviated notationsrepresenting concentrated and diluted, as follows:EMeone Veone Mail VailBased on the inverse nature of this equivalency, small volumes of concentrated solutions can be used to prepare large volumes of diluted solutions with a known concentration, which isadvantageous in areas involving solution chemistry.Part AIn order to understand how this equation is derived and why it holds true, the product of each side of the equation should be examined. What value and unit do you get when youmultiply a concentration of 0.735 M by a volume of 0.400 L ? This question can be expressed asExpress your answer to three significant figures with the appropriate units.► View Available Hint(s)M₁ V₁ = ValueSubmitPart B80$4RμAF4%5openvellum.ecollege.com→UnitsFST6?MacBook AirF6Y0.400 L X 0.735 mol =1&7P PearsonF7U8DIIFB(9DDF90Review I Constants I Periodic Table)0F10ⒸP-CF11++=F1288 BM₁V₁= ValueSubmitPart BWhen you need to produce a variety of diluted solutions of a solute, you can dilute a series of stock solutions. A stock solution has a significantly higher concentration of the givensolute (typically 10¹ to 10 times higher than those of the diluted solutions). The high concentration allows many diluted solutions to be prepared using minimal amounts of the stocksolution.What volume of a 6.73 M stock solution do you need to prepare 100. mL of a 0.2100 M solution of KCI?Express the volume to three significant figures with the appropriate units.► View Available Hint(s)V₁ = ValueSubmit80$HÅ4ER1F4UnitsDilution as a procedure in analysisWe sometimes purposely dilute solutions if the analyte concentrations are too high to permit an accurate measurement. For example, a spectrophotometer can measure the amount oflight that passes through a sample solution for wavelengths in the visible (and sometimes ultraviolet) range. Based on the dissolved solute's properties, we can quantify that solute'sconcentration. If the solute's concentration is too high, a relatively small amount of incident light having the wavelength that corresponds to the solute's absorption behavior will reach thedetector, which results in an inaccurate measurement (consider the result if you blocked the light with an opaque disc). If the solution is diluted, less light is absorbed, and the detector inthe spectrophotometer can generate more reliable data.When talking about small quantities, it is important to understand the relationship between common prefixes. Here are some relationships that may help when working through Part C.1L 1000 mL== 1,000,000 μLOR1 mL =AnP Pearson5UnitsFSopenvellum.ecollege.comDWGw ?6F60.001 LMacBook AirTY=&744F7U8DIIF81000 μL(9F90)0AF10-PⒸ +F11+=F128

Question

Accepted Answer

Answered: Image /qna-images/answer/cacae2b4-99d6-493c-8671-1e5bd69f6879.jpg