A rock salt (NaCl), ice, and water mixture is used to cool milk and cream to make homemade ice cream. How many grams of rock salt must be added to water to lower the freezing point 16.6°C?
\[\Delta T_f=K_b*m*i\]\[-16.6K=(1.853\frac{K*kg}{mol})*(\frac{mol(NaCl)}{kg(Sol'n)})*2\]\[4.479\frac{mol}{K*kg}=\frac{mol_{NaCl}}{kg_{Sol'n}}\]From here, no matter what type of simplification I do, I can't seem to eliminate it down to one variable. I think there may not be enough information. I'll keep trying, though.
This is my best guess, but I'm unsure that it's correct:\[\frac{\Delta T_f}{K_b*i}=m\]\[\frac{-16.6K}{(-1.853\frac{K*kg}{mol})(2)}=m\]\[m=4.4792\frac{kg}{mol}=4.4792m\]\[m_{NaCl}=\frac{mols(NaCl)}{kg}\]\[4.4792m=\frac{mols(NaCl)}{1kg}\]\[mols(NaCl)=4.4792mol\]\[4.4792mol(NaCl)*\frac{58.4427g(NaCl)}{1mol(NaCl)}=262g(NaCl)\]Once again, I'm not sure about this answer. I'd feel better if someone who knew exactly what they were doing checked it.
looks good to me, the only thing missing is the mass of water used, but if we assume 1.00kg, then the numbers work out
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