CHEM 1011 Quiz 2 Questions With Complete Solutions

What concentration of phosphate would give you a percent transmittance of 35%? correct answer: To find this we turn %T into an absorbance value, then we plug into the equation A = k * c. We know k from the slope of our line (in this case we will use the sample given 1245) and we calculate A from the %T. We then divide to find C, which is our concentration aka molarity. In this case it comes out to 3.7 x 10⁻⁴ = M. A worker dumps 20 grams of detergent that is 30% by weight phosphate ion into a 50 gallon (200L) aquarium. What would the concentration of the phosphate ion be? What would the absorbance be? Would the fish be in danger according to the EPA? correct answer: First we much find out how many moles of actual phosphate ion were dumped in. If 30% of the 20 grams dumped in was phosphate ion, that means 6 grams of phosphate ion was dumped in. Phosphate ion weights 96g/mol so we have 0.0625 moles of phosphate. Dumping this into the tank, we have .0625moles/200L which equates out to 3.13 x 10⁻⁴ M. For absorbance, we plug that into the A = kc equation. k = 1245 and c = 3.13 x 10⁻⁴ so that gives us A which is approx .393. Doing the math, we have an EPA defined limit of 3.16 x 10⁻⁶ M limit for phosphate concentration so the fish gonna die. You find that the standard curve for phosphate is A = 1245 M. a) What is the molarity of phosphate in a diluted unknown solution that has a %T of 27.1%? b) The dilute solution was made by taking 1.000 mL of unknown and diluting it to 15.00 mL. What is the molarity of the original unknown? correct answer: First we turn 27.1%T into Absorbance, which ends up being 0.567. Plugging that along with 1245 = k into the A = kc equation, we get that c = 4.55 x 10⁻⁴ M. To find the molarity of the original equation, we do M₁V₁ = M₂V₂. This yields 4.55 x 10⁻⁴ * 15mL / 1m