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A $4~\mathrm{g}$ mixture of sucrose ($\ce{C12H22O11}$) and zinc nitrate ($\ce{Zn(NO3)2}$) is dissolved in $150~\mathrm{g}$ of water. If the resulting solution freezes at $-0.768~^\circ\mathrm{C}$, what is the mass percent of sucrose in the mixture? Assume that zinc nitrate dissociates completely in water, $K_\mathrm f = 1.86$. (Answer: $2.8\%$)

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First find the molality of solute particles. Assume i = 1; we're just counting how many things are dissolved and it doesn't matter whether they are ions or molecules at this point. I'm using m for mass and b for molality throughout to avoid confusion.

$\Delta T = ibk_f\implies\frac{\Delta T}{k}=b$

$b=\frac{0.768}{1.86}=\pu{0.413 molal}$

Next find how many moles of particles were dissolved.

$b = \frac{n_{\text{solute}}}{m_{\text{solvent}}}$

$n_{\text{solute}}=b\times m_{\text{solvent}}=0.413\times 0.150 = 0.0619~\text{moles of particles}$

Now determine the mass of each species based on this total number of moles.
$n_{\text{particles}}=3n_{\ce{Zn(NO3)2}}+n_{\ce{C12H22O11}}$

$n_{\text{particles}}=3\frac{m_{\ce{Zn(NO3)2}}}{M_{\ce{Zn(NO3)2}}}+\frac{m_{\ce{C12H22O11}}}{M_{\ce{C12H22O11}}}$

You know the total number of moles of particles, 0.0619, and you know the molar masses of both zinc nitrate and sucrose, but you don't know either the actual mass of zinc nitrate or of sucrose. However, you do have one other piece of information that allows you to set up a system of equations.

$m_{\text{solute}}=m_{\ce{Zn(NO3)2}}+m_{\ce{C12H22O11}}=\pu{4g}$

Solving this system of equations should give you the masses of both solutes.

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Hint: $$i\neq3\text{, but }\underbrace{\frac x{342}}_{\text{sucrose}}+\underbrace{\frac{4-x}{189}\times3}_{\text{zinc-nitrate}}$$

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