At –12.5 °C (a common temperature for household freezers), what is the maximum mass of saccharin...

At –12.5 °C (a common temperature for household freezers), what is the maximum mass of saccharin (C7H5NO3S) you can add to 1.00 kg of pure water and still have the solution freeze? Assume that saccharin is a molecular solid and does not ionize when it dissolves in water. Kf values are given here.

Solvent	Formula	K_f value* (°C/m)	Normal freezing point (°C)	K_b 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

Expert Solution

We know the equation for the freezing point depression is:

T = k_f*m

where,

T is the depression in the freezing point

k_f is the molal freezing point depression constant

m is the molality of the solute i.e. moles per kg of a solute

Given T = 12.5^oC, k_f = 1.86

Therefore molality m = T/k_f = 12.5/1.86 = 6.72 moles saccharin per kg of water

This also is equal to 6.72 * MM_saccharin = 6.72*183.18 = 1231g Saccharin per kg water.

Below this amount the water will still freeze at the given temperature.

queen_honey_blossom answered 2 years ago

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