Colligative properties : Consider these two solutions: Solution A is prepared by

Question

Colligative properties :

Consider these two solutions: Solution A is prepared by dissolving 5.00 g of MgCl, in enough water to make 0.250 L of solution, and Solution B is prepared by dissolving 5.00 g of KCl in enough water to make 0.250 L of solution. Which direction will solvent initially flow if these two solutions are separated by a semipermeable membrane? 12. Assuming that the volumes of the solutions described in question #12 are additive and ignoring any effects that gravity may have on the osmotic pressure of the solutions, what will be the final volume of solution A when the net solvent flow through the semipermeable membrane stops? 13. a+ 2 ey

Explanation / Answer

12)

step 1: find osmotic pressure of MgCl2

Molar mass of MgCl2 = 1*MM(Mg) + 2*MM(Cl)

= 1*24.31 + 2*35.45

= 95.21 g/mol

mass of MgCl2 = 5.00 g

we have below equation to be used:

number of mol of MgCl2,

n = mass of MgCl2/molar mass of MgCl2

=(5.0 g)/(95.21 g/mol)

= 5.252*10^-2 mol

volume , V = 0.250 L

we have below equation to be used:

Molarity,

M = number of mol / volume in L

= 5.252*10^-2/0.25

= 0.2101 M

T = 298 K

we have below equation to be used:

P = i*C*R*T

P = 3.0* 0.2101*0.0821*298.0

P =15.4 atm

step 2: find osmotic pressure of KCl

Molar mass of KCl = 1*MM(K) + 1*MM(Cl)

= 1*39.1 + 1*35.45

= 74.55 g/mol

mass of KCl = 5.00 g

we have below equation to be used:

number of mol of KCl,

n = mass of KCl/molar mass of KCl

=(5.0 g)/(74.55 g/mol)

= 6.707*10^-2 mol

volume , V = 0.250 L

we have below equation to be used:

Molarity,

M = number of mol / volume in L

= 6.707*10^-2/0.25

= 0.2683 M

T = 298 K

we have below equation to be used:

P = i*C*R*T

P = 2.0*0.26827632461435275*0.0821*298.0

P =13.1 atm

Since osmotic pressure is more for MgCl2, solvent will flow to the side of MgCl2

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