Test Index

Solutions Part 2

Section: Chemistry

Question : 57 of 95

Marks: +1, -0

Solution:

\Delta T_{f}

is maximum when

i \times m

is maximum.

m_1 = \frac{180}{60} = 3, i = 1+\alpha

Hence

\Delta T_f = (1+\alpha) \cdot k_f = 3 \times 1.86

= 5.58^{\circ} \mathrm{C} (\alpha << 1)

m_2 = \frac{180}{60} = 3, i = 0.5, \Delta T_f

= \frac{3}{2} \times k_f' = 7.68^{\circ} \mathrm{C}

m_3 = \frac{180}{122} = 1.48, i = 0.5,

\Delta T_f = \frac{1.48}{2} \times k_f' = 3.8^{\circ} \mathrm{C}

m_4 = \frac{180}{180} = 1, i = 1,

\Delta T_f = 1 \cdot k_f' = 1.86^{\circ} \mathrm{C}

As per NCERT,

k_f' (\mathrm{H}_2\mathrm{O})

= 1.86 \mathrm{K} \cdot \mathrm{kg} \mathrm{mol}^{-1}

k_f' (\text{Benzene}) = 5.12 \mathrm{K} \cdot \mathrm{kg} \mathrm{mol}^{-1}

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