Answer:
[tex]\large \boxed{\text{c = 2.50 mol/L; b = 3.96 mol/kg }}[/tex]
Explanation:
1. Molar concentration
Let's call chloroform C and acetone A.
Molar concentration of C = Moles of C/Litres of solution
(a) Moles of C
Assume 0.187 mol of C.
That takes care of that.
(b) Litres of solution
Then we have 0.813 mol of A.
(i) Mass of each component
[tex]\text{Mass of C} = \text{0.187 mol C} \times \dfrac{\text{119.38 g C}}{\text{1 mol C}} = \text{22.32 g C}\\\\\text{Mass of A} = \text{0.813 mol A} \times \dfrac{\text{58.08 g A}}{\text{1 mol A}} = \text{47.22 g A}[/tex]
(ii) Volume of each component
[tex]\text{Vol. of C} = \text{22.32 g C} \times \dfrac{\text{1 mL C}}{\text{1.48 g C}} = \text{15.08 mL C}\\\\\text{Vol. of A} = \text{47.22 g A} \times \dfrac{\text{1 mL A}}{\text{0.791 g A}} = \text{59.70 mL A}[/tex]
(iii) Volume of solution
If there is no change of volume on mixing.
V = 15.08 mL + 59.70 mL = 74.78 mL
(c) Molar concentration of C
[tex]c = \dfrac{\text{0.187 mol}}{\text{0.07478 L}} = \textbf{2.50 mol/L }\\\\\text{ The molar concentration of chloroform is $\large \boxed{\textbf{2.50 mol/L}}$}[/tex]
2. Molal concentration of C
Molal concentration = moles of solute/kilograms of solvent
Moles of C = 0.187 mol
Mass of A = 47.22 g = 0.047 22 kg
[tex]\text{b} = \dfrac{\text{0.187 mol}}{\text{0.047 22 kg}} = \textbf{3.96 mol/kg }\\\\\text{The molal concentration of chloroform is $\large \boxed{\textbf{3.96 mol/kg}}$}[/tex]
