Question

In the lab, you weigh an empty beaker and find it to in empty beaker and find it to be 28.3456 grams. You use a volumetric pipet to transfer 50.00 ml of a sodium chloride solution in beaker and weigh the beaker again to get 80.5678 grams. You being careful to turn down the heat near the end of the process minimize the amount of 'popping' that occurs. At the v mount of 'popping' that occurs. At the very end, you dry the beaker in the oven to make sure the salt is completely dry. After cooling the beaker, you weigh it again and find it to be 29.1234 grams. in to get 80.5678 grams. You boil the salt water, te heat near the end of the process in order to Calculate the m/v % of the original solution: Calculate the m/m % of the original solution: Calculate the molarity of the original solution: Give one possible reason your answer would be higher than the actual value: Give one possible reason your answer would be lower than the actual value:

Answer 1

Mass of beaker = 28.34569 cemply) = 28.3456g Volume of solution = 50mL mass of beaker & - 80.5678 g mass of solution = 80.5678 - 28.3456 g

UWDWW - 52.22229 mass of beaker + salt = 29.1234g mass of salt = 29.1234g - 28.3456g 2 0.77 c) Molarity = mass of solute molar mass solute X Volume (L) - 0.7778 8 58.5 of motx 50NoL (Nacl = 58.5g/mol) a) We have mass of salt (solute) = 0.7776 g TM = 0.01328M I mass of solution = 52.222g d) Because everytime we heat the " " Solvent 52.2222-0. 7770 | beaker, the mass of weigh beaker does not show the exact weight in -51.4444 Volume=SomL the weighing Balance. This is because 50m2 of solection contains 0.77789 of crystalline nature of glass. Nach i loomL . . .. 77 78 x2) e) while evaporating solvent, due = 1.5556 g of to prolonged heating it might I cause bumping of solvent out - m/vs. = 1.5556%. an original which causes spill of solerte too dissolved is water to come out. I b) : 52.2222g of solution contains Thereky giring a low value solution Naci Solution 0.7778 & Nach : loog » 4 & 0.7778 g xloog 52.2222 g - 1.489 g mlms - 1.489). I