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So here is my question,

Question: A water sample contains 9.5% $\mathrm{MgCl_2}$ and 11.7% $\mathrm{NaCl}$ (by weight). Assuming 80% ionization of each salt. Boiling point of water will be ________. ($\mathrm{K_b}$ for water $\mathrm{=0.52}$)

You know, this question isn't that tough to bother anyone (definitely not!). What's all confusing here is the molality of two solutes.

I know that,

$$\Delta \mathrm{T_b=K_b\cdot m}$$

Degree of dissosiation $(\mathrm{\alpha}) = 0.8$

So I know very well that molality of a solution having only one solute is,

$$\mathrm{m=\frac{n\ (no.\ of\ moles\ of\ solute)}{y\ (mass\ of\ solvent\ in\ kg)}}$$

But what about the solution having more than one solute? Do they have any molality (I am sure that there will surely be, because it is there in the question). If yes then what's the method of calculating?

Also if you are solving this, here is the answer to the question (as per my answer key),

377 K

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The boiling point elevation fomula assumes that all particles are equal, only the number of particles in the solution matters.

So I would use:

m = (0.2 + (3)(0.8))(molality of $\ce{MgCl_2}$) + (0.2 + (2)(0.8))(molality of $\ce{NaCl}$)

To find molality of $\ce{MgCl_2}$ and molality of $\ce{NaCl}$, consider a 1kg sample of the solution. There are 95g $\ce{MgCl_2}$, 117g $\ce{NaCl}$ and 788g water.

Calculate the number of moles $\ce{MgCl_2}$ in 95g $\ce{MgCl_2}$, and divide this number by 0.788kg. This is the molality of $\ce{MgCl_2}$ in the solution.

Repeat for $\ce{NaCl}$

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