Question

For each of the following, provide the appropiate value:

a. What is the molar mass of a non-ionizing, non-volatile solid that when 14.2 g is dissolved in water to make 100.0 mL of solution creats a solution with an osmotic pressure of 5.17 atm at 25.0 ^oC

b. What is the freezing point of an aqueous solution made by dissolving 0.158 mole of MgCl₂ in 100.0 g of water? K_f = 1.86 ^oC/m

c. What is the total vapor pressure in atm of a solution created by combining 3.44 moles of toluene and 5.99 moles of benzene at 350K? At 350K P_{vap, toluene} = 0.33 atm and P_{vap, benzene} = 0.95 atm

Answer 1

a)

MW of

m = 14.2 g

V = 100 ml = 0. 1L

P = 5.17 atm at 25°C

PI = M*RT

5.17 = M*0.082*298

M = 5.17/(0.082*298) = 0.21157308888

M = mol/L

mol = M*L = 0.21157308888*0.1

= 0.021157 mol of substance

MW = mass/mol = 14.2/0.21157308888 = 67.116 g/mol

b)

dTF = -Kf*m*i

i = 3 ions Mg+2 and 2Cl-

m = 0.158 / (0.1)= 1.58

dTf = -1.86*1.58*3 = - 8.8164°C

c)

P° = xa*P°a + (1-x1)*P°b

P° = (3.44/(5.99 + 3.44 )) * 0.33 +   (5.99/(5.99 + 3.44 )) *0.95 =0.72382820784 atm