Please get full work. Thanks :D Solutions and Dilutions.Revisited (a) How many g
ID: 570815 • Letter: P
Question
Please get full work. Thanks :D
Solutions and Dilutions.Revisited (a) How many grams of CuSo 5 H20 should be dissolved in sufficient water to prepare 250.0 1. mL of solution that is 6.00 mM in Cu2? (b concentrated perchloric acid is commercially available as a 70.0 wt % aqueous solution and has a density of 1.67 gmL-1. What volume of this reagent should be diluted with water to make 500.0 mof 0.100 M HCIO4 (aq)? (c) A 150.0 mL portion of aqueous solution contains 3.172 g of potassium chromate. A 25.0 aliquot of the solution is diluted to a volume of 5.00 mL. What is the concentration of chromate ion in the diluted solution in g mL-17 what is the concentration of chromate ion in the diluted solution in mol L (d) In an environmental analysis project, the concentration of the herbicide dicamba (CsHsCl203) was determined in several samples of river water. Solid phase extraction (SPE) was used to "pre-concentrate" the sample prior to analysis. 500.0 mL of river water was extracted, and the analyte was eluted into 2.50 mL of the solvent ethanol. Then, 100.0 L aliquots of the ethanol solutions were analyzed using GC-MS. These ethanol solutions were found to contain a dicamba concentration of 65.3 ng mL-1. What is the concentration of dicamba (in mol L-1) in the original river water sample?Explanation / Answer
a) Molar mass of CuSO4.5H2O = [1*63.546 + 1*32.065 + 4*15.9994 + 5*2*1.008 + 5*15.9994] g/mol = 249.6856 g/mol.
CuSO4.5H2O ionizes as below.
CuSO4.5H2O (aq) --------> Cu2+ (aq) + SO42- (aq) + 5 H2O (l)
As per the stoichiometry of the reaction,
1 mole CuSO4.5H2O = 1 mole Cu2+.
Mole(s) of Cu2+ in 250.0 mL of 6.00 mM solution = (250.0 mL)*(1 L/1000 mL)*(6.00 mM)*(1 M/1000 mM)*(1 mol.L-1/1 M) = 0.0015 mole.
Mass of CuSO4.5H2O required to prepare the given solution = (0.0015 mole)*(249.6856 g/mol) = 0.3745284 g 0.3745 g (ans).
b) A 70.0 wt% aqueous solution of HClO4 contains 70.0 g HClO4 per 100.0 g water.
The density of the solution is 1.67 g.mL-1; hence, the weight of the aqueous solution is (70.0 + 100.0) g = 170.0 g and the volume of the aqueous solution = (170.0 g)/(1.67 g.mL-1) = 101.7964 mL.
Molar mass of HClO4 = (1*1.008 + 1*35.453 + 4*15.9994) g/mol = 100.4586 g/mol.
Mole(s) of HClO4 corresponding to 70.0 g HClO4 = (70.0 g)/(100.4586 g/mol) = 0.6968 mole.
Molar concentration of the HClO4 solution = (moles of HClO4)/(volume of solution in L) = (0.6968 mole)/[(101.7964 mL)*(1 L/1000 mL)] = 6.8450 mol/L 6.845 M.
Use the dilution equation:
M1*V1 = M2*V2 where M1 = 6.845 M; M2 = 0.100 M and V2 = 500.0 mL.
Plug in values and get
(6.845 M)*V1 = (0.100 M)*(500.0 mL)
====> V1 = 7.3046 mL 7.30 mL.
7.30 mL of the prepared HClO4 must be diluted to get the desired concentration.