Question

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

Answer 1

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