Assuming 100% dissociation, calculate the freezing point and boiling point of 1.52 m SnCl4(aq). Constants may be found here.
vent | Formula | Kf
value*
(°C/m) |
Normal freezing point (°C) |
Kb
value
(°C/m) |
Normal boiling point (°C) |
water | H2O | 1.86 | 0.00 | 0.512 | 100.00 |
benzene | C6H6 | 5.12 | 5.49 | 2.53 | 80.1 |
cyclohexane | C6H12 | 20.8 | 6.59 | 2.92 | 80.7 |
ethanol | C2H6O | 1.99 | –117.3 | 1.22 | 78.4 |
carbon tetrachloride |
CCl4 | 29.8 | –22.9 | 5.03 | 76.8 |
camphor | C10H16O | 37.8 | 176 |
I have solved the first four sub-parts, please post one more question to get the remaining answer
Vont hoff factor for SnCl4 (i) = 5
Depression in Freezing point of water= i * KF * m
=> 5 * 1.86 * 1.52
=> 14.136
Hence freezing point of water = -14.136C
Elevation in boiling point of water= i * Kb * m
=> 5 * 0.512 * 1.52
=> 3.8912
Hence boiling point of water = 103.8912 C
Depression in Freezing point of benzene= i * KF * m
=> 5 * 5.12 * 1.52
=> 38.912
Hence freezing point of benzene = -33.422C
Elevation in boiling point of benzene = i * Kb * m
=> 5 * 2.53 * 1.52
=> 19.228
Hence boiling point of benzene = 99.328 C
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