QuestionJuly 7, 2025

What is the freezing point, in {}^circ C of a 0.41 m solution of C_(5)H_(4) in benzene? FP_((benzene))=5.50^circ C i=1 K_(f(benzene))=5.12^circ C/m Delta T_(f)=(i)(K_(f))(m) [?]cdot ^circ C

What is the freezing point, in {}^circ C of a 0.41 m solution of C_(5)H_(4) in benzene? FP_((benzene))=5.50^circ C i=1 K_(f(benzene))=5.12^circ C/m Delta T_(f)=(i)(K_(f))(m) [?]cdot ^circ C
What is the freezing point, in {}^circ C of a
0.41 m solution of C_(5)H_(4) in benzene?
FP_((benzene))=5.50^circ C i=1
K_(f(benzene))=5.12^circ C/m
Delta T_(f)=(i)(K_(f))(m)
[?]cdot ^circ C

Solution
4.1(259 votes)

Answer

3.40^{\circ }C Explanation 1. Calculate the freezing point depression Use the formula \Delta T_{f} = (i)(K_{f})(m). Substitute i = 1, K_{f} = 5.12^{\circ }C/m, and m = 0.41 m to get \Delta T_{f} = (1)(5.12)(0.41). 2. Compute the value of \Delta T_{f} Calculate \Delta T_{f} = 5.12 \times 0.41 = 2.0992^{\circ }C. 3. Determine the new freezing point Subtract \Delta T_{f} from the normal freezing point of benzene: FP_{(benzene)} - \Delta T_{f} = 5.50^{\circ }C - 2.0992^{\circ }C.

Explanation

1. Calculate the freezing point depression<br /> Use the formula $\Delta T_{f} = (i)(K_{f})(m)$. Substitute $i = 1$, $K_{f} = 5.12^{\circ }C/m$, and $m = 0.41$ m to get $\Delta T_{f} = (1)(5.12)(0.41)$.<br /><br />2. Compute the value of $\Delta T_{f}$<br /> Calculate $\Delta T_{f} = 5.12 \times 0.41 = 2.0992^{\circ }C$.<br /><br />3. Determine the new freezing point<br /> Subtract $\Delta T_{f}$ from the normal freezing point of benzene: $FP_{(benzene)} - \Delta T_{f} = 5.50^{\circ }C - 2.0992^{\circ }C$.
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