First slide

Theory of dilute solutions

Question

An aqueous solution freezes at -0.186 $°$ C (K_f= 1.86 $°$ ; K_b= 0.512 $°$ ).What is the elevation in boiling point?

Moderate

A

0.186
B

0.512
C

$\frac{0.512}{1.86}$
D

0.0512

Solution

$∆ T_{b} = K_{b} \times M o l a l i t y ∆ T_{f} = K_{f} \times M o l a l i t y ∴ \frac{∆ T_{b}}{∆ T_{f}} = \frac{K_{b}}{K_{f}} \Rightarrow ∆ T_{b} = \frac{∆ T_{f} \times K_{b}}{K_{f}} = \frac{0.186 \times 0.512}{1.86} = 0.0512$

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