An aqueous solution has a normal boiling point of 103.0

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Question

Answer 1

Answer #1

$\Delta T_b=k_b*m\ \\ \\ m=\frac{\Delta T_b}{k_b}=\frac{3}{0.520}=5.76\ \\ \\ \Delta T_f=k_f*m=1.86*5.76=10.71\approx 11C$

Thus the freezing point is -11C

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Answer 2

Answer #2

I think that -11

Answer 3

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