The limiting molar conductivities of the following salts were measured in methan

Question

The limiting molar conductivities of the following salts were measured in methanol solvent: KN03 (114.5); KCI (105.0); LiCI (90.9) (all In S cm^2 mol^-1.) Calculate A^degree for LiN03 in methanol. A conductivity cell had electrodes 2.00 cm* In area separated by 1.00 cm. When filled with a solution containing 50.0 g KCI(s) dissolved in a liter of aqueous solution, the resistance was 7.25 ohms. Estimate the molar conductivity of KCI under these conditions. A different conductivity cell filled with 0.100 molar KCI(aq) at a certain temperature shows a resistance of 24.96 ohms. This solution at this temperature has a known conductivity of 0.011639 ohm^-1 cm^-1. The conductivity of water used to make solutions for use in this cell was 7.5 times 10^-8 ohm^-1 cm^-1, When the cell was filled with 0.0100 molar acetic add (aq), the measured resistance was 1982 ohms, (a) Calculate the cell constant, (b) Calculate the molar conductivity of acetic acid under these conditions. The infinitely dilute molar conductivity of the hydrogen ion and the acetate ion are 349.8 and 40.9 ohm^-1 cm^2 mole^-1 respectively, (a) Calculate alpha (the fraction of dissociation) of acetic acid if a 1,00 times 10^-6 M nominal concentration of aqueous acetic acid had a conductivity of 4.46 times 10^-7 ohm^-1 cm^-1 when the water used to make the solution had a conductivity of 7.5 times 10^-8 ohm^-1 cm^-1, (b) From your alpha value, calculate Ka, the acid dissociation value of acetic acid. The infinitely dilute molar conductivity of the silver ion and the chloride ion are 56.9 and 68.4 ohm^-1 cm^2 mole^-1 respectively. A saturated aqueous solution of silver chloride has a conductivity of 1.33 times 10^-6 ohm^-1 cm^-1, (a) calculate the solubility of silver chloride in units of molarity, (b) calculate the Ksp of silver chloride.

Explanation / Answer

Q1) The limiting molar conductance of LINO3 = conductance of [LiCl + KNo3] - condctance ofKCl according to Kolhrausch law.

thus conductance of LiNO3 = 114.5 + 90.9 -105.0

= 100.4

2) From the given data cell constant = l/a = 1/2

resistance R = 7.25 ohms

molarity of KCl ==50g/74.5 g/mol

= 0.67 M

We have conductivity k = (1/R) x cell constant

= (1/7.25) x 0.5

We know molar conductance = conductivity x 1000/Molarity

=[(1/7.25) x 0.5 x1000] / 0.67

   = 102.93 S Cm2 /mol

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