1) Which of the following aqueous solutions has the highest boiling point (assum

Question

1) Which of the following aqueous solutions has the highest boiling point (assume 100% dissociation for all soluble ionic compounds)? A)0.10m AI(NO B) 0.11m Na:So. Co.15m K:CO, D)0.18m NaCI ) 0.35m CoH Answer: ( 2) Assuming 00% dissociation, which of the following compounds is listed correctly with its van't Hoff factor? Answer 3) Calculate the molalities of the following aqucous solutions: (a) 1.22 M glucose (CHi0%) solution (density of solution 1.12 g/mL), (b) 0.87 M NaOH solution (density of solution -1.04 g/mL.)(c) 5.24 MNaHCOs solution (density of solution -1.19 g/mL) o Specify what ions are present upon dissolving cach of the following substances in water: (a) Mgla. (b) AINO,) (c) HCIO4, (d) NaCH,Coo. freezing-point depression to find the molar mass of a compound, a solution of 3.80 g of 5) In the experiment of using a compound molar mass of the solute and its molecular formula. (For benzene, the normal freezing point is 5.5°C and Kr value is having the empirical formula CoHsP in 19.6 g of benzene is observed to freeze at 3.2°C. Calculate the 5.12°C/m.)

Explanation / Answer

Ans. #3.A. 1.22 M glucose.

1.22 M solution consists of 1.22 mol glucose per L of solution.

Let the volume of solution be 1.0 L. So, moles of glucose in it = 1.22 mol.

# Mass of glucose = Moles x Molar mass = 1.22 mol x (180.15768 g/ mol)

= 219.792 g

# Mass of solution = Volume x Density = 1000.0 mL x (1.12 g/ mL) = 1120.0 g

# Mass of solvent (water) = Mass of solution – Mass of glucose

                                    = 1120.0 g – 219.972 g

                                    = 900.208 g

                                    = 0.900208 kg

# Now, molality = Moles of glucose / Mass of solvent in kg

                                    = 1.22 mol / 0.900208 kg

                                    = 1.355 m

#3.B. 0.87 M NaOH.

Let the volume of solution be 1.0 L. So, moles of NaOH in it = 0.87 mol.

# Mass of NaOH = 0.87 mol x (40.0 g/ mol) = 34.8 g

# Mass of solution = Volume x Density = 1000.0 mL x (1.04 g/ mL) = 1040.0 g

# Mass of solvent (water) = Mass of solution – Mass of NaOH

                                    = 1040.0 g – 34.8 g

                                    = 1005.2 g

                                    = 1.0052 kg

# Now, molality = 0.87 mol / 1.0052 kg = 0.865 m

#3.C. 5.24 M NaHCO3

Let the volume of solution be 1.0 L. So, moles of NaHCO3 in it = 5.24 mol.

# Mass of NaHCO3 = 5.24 mol x (84.0 g/ mol) = 440.16 g

# Mass of solution = Volume x Density = 1000.0 mL x (1.19 g/ mL) = 1190.0 g

# Mass of solvent (water) = Mass of solution – Mass of NaHCO3

                                    = 1190.0 g – 440.16 g

                                    = 749.84 g

                                    = 0.74984 kg

# Now, molality = 5.24 mol / 0.74984 kg = 6.988 m

           

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