3) Calculate the molalities of the following aqueous solutions: (a) 1.22 solutio

Question

3) Calculate the molalities of the following aqueous solutions: (a) 1.22 solution-1 .12 g/mL), (b) 0.87 M NaOH solution (density of solution = 1 04 g/mL), (c) 5.24 M NaHCO3 solution M glucose (CoH120s) solution (density of (density of solution = 1.19 g/mL). 4) Specify what ions are present upon dissolving each of the following substances in water: (a) Mgl, ()AINO>), (c) HCIO4, (d) NaCH3COO. 5) In the experiment of using freezing-point depression to find the molar mass of a compound, a solution of 3.80 g of a compound having the empirical formula CsHsP in 19.6 g of benzene is observed to freeze at 3.2°C. Calculate the molar mass of the solute and its molecular formula. (For benzene, the normal freezing point is 5.5°C and Kr value is 5.12°C/m.)

Explanation / Answer

Molarity = moles of solute/ liter of solution and molality= moles of solute/ kg of solvent

Basis: 1 liter of solution, 1000ml= 1L, volume of solution = 1000 ml, density = 1.12 g/ml, mass of the solution = 1000*1.12= 1120 gm, molarity = 1.22 M= 1.22 moles of glucose in 1L of solution. Since the basis is 1L, moles of glucose = 1.22 moles, mass of glucose= moles* molar mass = 1.22*180 gm =219.6 gm, mass of water= mass of solution-mass of glucose = 1120-219.6 =900.4 gm, 900.4/1000kg= 0.9004 kg, molality= 1.22/0.9004=1.35m

2. moles of NaOH with a basis of 1 Liter (1000ml ) of solution =0.87 moles, mass of NaOH= moles* molar mass=0.87*40 =34.8 gm, mass of solution = 1000*1.04 gm= 1040gm, mass of water= 1040-34.8 =1005.2 gm= 1005.2/1000 kg = 1.0052 kg, molality= 0.87/1.0052=0.865

3. moles of NaHCO3 in 5.24 moles with a basis of 1 liter of solution = 5.24 moles, mass of NaHCO3= 5.24*84 gm =440.16 gm, mass of solution = 1000*1.19 = 1190 gm, mass of water= 1190-440.16 =749.84 gm= 0.74984 kg, molality= 5.24/0.74984=7m

MgI2 ionizes to give Mg+2 and I-, Al(NO3)3 gives Al+3 and NO3- and HClO4 gives H+ and ClO4-, while CH3COONa gives Na+ and CH3COO-.

Freezing point depression =5.5-3.2= 2.3 deg.c

Freezing point depression = i*kf*m, i= Van;t Hoff factor= 1 for benzene and m= molality= moles of compound/ kg of benzene , moles = mass/molar mass , let M= Molar mass of compound

Moles of compound = 3.8/M, mass of benzene in kgs= 19.6/1000 kg =0.0196 kg

Molality= 3.8/(M*0.0196)

Hence 2.3= 1*5.12*3.8/(M*0.0196)

M= 5.12*3.8/(2.3*0.0196)= 432 g/mole

Given the empirical formula is C6H5P, molar mass = 6*12+5*1+31= 108 g/mole,

Molar mass of empirical formula*n= 432, n= no of molar mass units

108n= 432, n=4

So the formula of compound is (C6H5P)4= C24H10P4

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