PART A Question 1: 10.0 mL red dye mixed with deionized water to make 250.0 mL solution %(v/v) concentration of red dye solution = %weight (%wt. = m/m) = mass of solutemass of solution x 100%(v/v) concentration of red dye solution = = 10.0 mL250.0 mL x 100 = 4.0 %(v/v) Question 2: Rubbing alcohol is 70% (v/v) isopropyl alcohol missed with water Rubbing alcohol = 70% (v/v) isopropyl alcohol + 30% water Therefore 70% Rubbing alcohol (64.0mL) = isopropyl alcohol = 70100 x 64.0mL=44.8mLHence 44.8 milliliters of isopropyl alcohol was used. Question 3: 0.84g of Mg (NO3)2 solution to prepare 500.0 mL (density = 1.0g/mL) %weight (%wt. = m/m) = mass of solutemass of solution x 100Mass of solute = 0.84g; Mass of solution = 500.0 mL = 500 mL x 1.0g/mL = 500g m/m = 0.84g500g x 100 = 0.168 Question 4: grams of CuSO4 to prepare of K2O with 0.048 g From Moles =mass of K2OMolar mass of K2O mass of K2O = moles of K2O x molar mass of K2O Molar mass of K2O: (K = 39 O = 16) = (39 x 2 + 16) = 94g From Molality m = moles of soluteMass of solvent in Kgs mass of solvent in Kg = moles of soluteMolality First the moles in K2O are obtained as follows: Moles =mass of K2OMolar mass of K2O = 0.048g94g = 5.106 x 10-4 mol. Mass of solvent is then computed as follows: Mass of solvent in Kg = moles of soluteMolality = 5.106 x 10 ^-4g0.03m = 0.01702g. Question 19: Dilution of CuCl2 Mc*Vc = Md*VdLet new concentration be y 2.0M x 10.0mL = 1000*y mL y = 2.0 x 10.01000 = 0.02M Question 20: Diluting 3.0M HNO3 New diluted HNO3 = 3.0 x 150 = 0.06M [...]
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