Question

This question has two parts (10 pts): (a) Consider the freezing point of 0.01M KCl(aq), tf(KCl), compared with that of 0.01 M Ca(NO3)2(aq), tf(Ca(NO3)2. Which of the following is true (circle one only)? i. tf(KCl) > tf(Ca(NO3)2) ii. tf(KCl) < tf(Ca(NO3)2) iii. tf(KCl) = tf(Ca(NO3)2) (b) What volume (in mL) of phenol (Mm = 94.1 g/mol; nonelectrolyte; density = 1.07 g/mL) must be added to 25.0 g of benzene(l) to have the freezing point of the resulting solution be ?0.80ºC? The kf (benzene) = 5.1 ºC/m. The freezing point of pure benzene is 5.5ºC.

Answer 1

a)

KCl dissociates as

KCl $\rightarrow$ K⁺ + Cl^- .

Hence, number of ions = 2.

As it is completely dissociates, then vant hoff factor (i) = 2.

Again Ca(NO₃)₂ being strong electrolyte dissociates as

Ca(NO₃)₂$\rightarrow$ Ca²⁺ + 2NO₃^-

So, number of ions = 3, then i = 3.

Now, freezing point depression;  $\Delta$T_f = i* K_f m.

So, 0.01 M Ca(NO₃)^{​​​​​​}₂ has higher value in depression of freezing point than 0.01 M KCl.

Therefore, 0.01 M Ca(NO₃)₂ has lower freezing point than 0.01 M KCl.

So, t_f(KCl)> t_f(CaNO₃)₂ .

Option i is correct.

B)

By Roult's law for nonelectrolyte

$\Delta$T_f = K_f*m

Or, $\Delta$T_f = K_f *
_{W2* 1000 MW LI}

W₂ is mass of solute (phenol)

M is molar mass of solute = 94.1 g/mol.

W₁ is mass of solvent ( benzene) = 25.0 g.

Now, given freezing point depression = $\Delta T$_f  = (T_f - T_f^' )

= (5.5 - 0.80) = 4.7⁰c .

4.7 =

5.1 W2 1000 94.1 25.0

Or, W₂ = (4.7*94.1*25.0)/(5.1*1000) =  2.168 g.

Given, Density of phenol = 1.07 g/mL.

Then , Volume of phenol = (2.168/1.07) = 2.026 mL.