(1)An aqueous solution contains 167g CuSO4 in 820mL ofsolution, the density of t
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Question
(1)An aqueous solution contains 167g CuSO4 in 820mL ofsolution, the density of the solution is 1.195g/mL calculate themolarity of the solution. (2) What is the molarity of a solution prepared by dissolving25.2g CaCO3 in 600mL of a water solution? (3) Calculate the molarity and molality of a solution thatcontains 18.0 percent HCl by mass and has a density of1.05g/ml (4) what is the molality of a 2.50 M aqueous solution ofethanol, C2H5OH ( molar mass = 46g/mol)?with density0.789g/ml? (1)An aqueous solution contains 167g CuSO4 in 820mL ofsolution, the density of the solution is 1.195g/mL calculate themolarity of the solution. (2) What is the molarity of a solution prepared by dissolving25.2g CaCO3 in 600mL of a water solution? (3) Calculate the molarity and molality of a solution thatcontains 18.0 percent HCl by mass and has a density of1.05g/ml (4) what is the molality of a 2.50 M aqueous solution ofethanol, C2H5OH ( molar mass = 46g/mol)?with density0.789g/ml?Explanation / Answer
QUESTION ONE n(CuSO4) = m/M = 167/159.61 = 1.046mol c(CuSO4) = n/v = 1.046/0.82 = 1.28M QUESTION TWO n(CaCO3) = m/M = 25.2/100.9 = 0.2518mol c(CaCO3) = n/v = 0.2518/0.6 =0.4196M QUESTION THREE Assume the volume to be 100mL Density = mass/volume 1.05 = mass/1000 Mass of solution= 1050g Mass of HCL (solute) = 1050 x 18% = 189g n(HCl) = m/M = 189/36.458 = 5.184mol Mass of solvent = 1050 - 189 = 861g Molarity = n/v = 5.184/1 = 5.184M Molality = 5.184/0.861 = 6.02m QUESTION FOUR Assume the volume to be 1000mL Density = mass/volume 0.789 = mass/1000 Mass of solution = 789g c(ethanol) = n/v n(ethanol) = cv = 2.5 x 1 = 2.5mol n(ethanol) = m/M m(ethanol) = nM = 2.5 x 46 = 115g Mass of solvent = 789 - 115 = 674g = 0.674kg Molality = 2.5/0.674 = 3.71m Hope this helps!
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