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enyata [817]
3 years ago
8

Can someone please help with this? ;)

Chemistry
1 answer:
Anvisha [2.4K]3 years ago
3 0

Answer:

The correct answer is A

Explanation:

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(3.20x10^4)(0.2402) express your answer in appropriate significant figures.
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A solution is made by dissolving 0.656 mol of nonelectrolyte solute in 869 g of benzene. calculate the freezing point, tf, and b
solmaris [256]
Answer is: the freezing point is 1.63°C and boiling point is 82.01°C.<span>.

1) n(</span><span>nonelectrolyte solute) = 0.656 mol.
</span>m(C₆H₆ - benzene) = 869 g ÷ 1000 g/kg.
m(C₆H₆) = 0.869 kg.<span>
b(solution) = n(</span>nonelectrolyte solute) ÷ m(C₆H₆).<span>
b(solution) = 0.656 mol ÷ 0.869 kg.
b(solution) = 0.754 mol/kg.

2) ΔT = Kf(benzene) · b(solution).
ΔT = 5.12°C/m · 0.754 m.
ΔT = 3.865°C.
Tf = 5.50°C - 3.865°C.
Tf = 1.63°C.
</span>
3) ΔTb = Kb(benzene) · b(solution).
ΔTb = 2.53°C/m · 0.754 m.
ΔTb = 1.91°C.
Tb = 80.1°C + 1.91°C.
Tb = 82.01°C.<span>

</span>
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What is the usual charge on an ion from group 7a
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