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bogdanovich [222]
3 years ago
13

After you preform your experiment, you determine that the kf value for naphthalene is 6.91 °c/m . you are using 10g of naphthal

ene and added 1.0 g of your unknown. the the freezing point of the solvent decreased by 4.47 °c when the unknown was added. knowing this information, determine the molar mass of the unknown. g/mol
Chemistry
1 answer:
Masteriza [31]3 years ago
8 0
Answer is: molar mass of compound is 154,58 g/mol.<span>
m(</span>naphthalene<span>) = 10 g = 0,01 kg.
m(unknown compound) = 1,00 g.
</span>Δ<span>T (solution) = 4,47 °C.
Kf(</span>naphthalene) = 6,91°C/m<span>; cryoscopic constant.
M</span>(unknown compound) = Kf(naphthalene)· m(unknown compound) ÷ m(naphthalene)<span> · ΔT(solution).
M(xylene) = </span>6,91°C/m<span> · 1 g ÷ 0,01 kg · 4,47</span>°C<span>.
M(xylene) = 154,58 g/mol.</span>
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mol\text{ }Li_3PO_4=3.40\text{ }mol\text{ }Ca(NO_3)_2*\frac{2\text{ }mol\text{ }Li_3PO_4}{3\text{ }mol\text{ }Ca(NO_3)_2}=2.2667\text{ }mol\text{ }Li_3PO_4

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4) Moles of Ca3(PO4)2 produced from the limiting reactant.

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The molar ratio between Ca(NO3)2 and Ca3(PO4)2 is 3 mol Ca(NO3)2: 1 mol Ca3(PO4)2.

mol\text{ }Ca_3(PO_4)_2=3.40\text{ }mol\text{ }Ca(NO_3)_2*\frac{1\text{ }mol\text{ }Ca_3(PO_4)_2}{3\text{ }mol\text{ }Ca(NO_3)_2}=1.1313\text{ }mol\text{ }Ca_3(PO_4)_2

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g\text{ }Ca_3(PO_4)_2=1.1333\text{ }mol\text{ }Ca_3(PO_4)_2*\frac{310.1767\text{ }g\text{ }Ca_3(PO_4)_2}{1\text{ }mol\text{ }Ca_3(PO_4)_2}g\text{ }Ca_3(PO_4)_2=351.526\text{ }g\text{ }Ca_3(PO_4)_2

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