Problem 1

An unknown salt

\ce{MX2}

is a group 2 metal halide. (a) 10.00 g of

\ce{MX2}

dissolves in 50.0 g of water to give a homogeneous solution whose freezing point is

-4.50\ ^{\circ}\mathrm{C}

. What is the molar mass of

\ce{MX2}

? For water,

K_f = 1.86\ ^{\circ}\mathrm{C}\,m^{-1}

. (b) 10.00 g of

\ce{Na2CO3}

and 10.00 g of

\ce{MX2}

are mixed in 200.0 mL of water and a precipitate of

\ce{MCO3}

forms. What is the pH of the supernatant? The

K_a

\ce{H2CO3}

4.3 \times 10^{-7}

and the

K_a

\ce{HCO3-}

4.7 \times 10^{-11}

. (c) A solution of 10.00 g of

\ce{MX2}

in water is treated with excess silver nitrate; the dried precipitate has mass 15.2 g. What is the identity of

\ce{MX2}

? (d) A sample of 10.00 g of

\ce{MX2}

dissolved in 50 mL of water is treated with increasing amounts of

\ce{Na2SO4}

up to 10 g in total; describe how the mass of precipitate varies with the mass of added

\ce{Na2SO4}

. (e) What colour flame test does

\ce{MX2}

give?

Step 1 of 4: Molar mass from freezing point

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Intuition

Colligative properties count particles: the van 't Hoff factor i = 3 halves-or-thirds the inferred molar mass.

\Delta T_f = i K_f m \Rightarrow m = \frac{4.50}{3 \times 1.86} = 0.806\ \text{mol kg}^{-1};\quad n = 0.0403\ \text{mol} \Rightarrow M(\ce{MX2}) = \frac{10.00}{0.0403} = 248\ \text{g mol}^{-1}

Analysis

A group-2 halide MX2 dissociates into three ions, so with $i = 3$ : molality $m = 4.50/(3 \times 1.86) = 0.806$ mol kg $^{-1}$ , i.e. 0.0403 mol in 50.0 g of water, giving $M = 10.00/0.0403 = 248$ g mol $^{-1}$ .