What is the freezing point of an aqueous solution that boils at 105.4 ∘c?

Answer : -19.6

Explanation : We can calculate this using the combine equation of boiling point and depression in freezing point and obtain a combined equation which is

∆T(f) = ∆T(b) x {K(f) / K(b)} ;

Where ∆T(f) is depression in freezing point,

∆T(b) is elevation in boiling point,

K(f) is cryoscopic constant for water it is 1.86 

K(b) is ebullioscopic constant for water it is 0.512

So we can solve the equation,

∆T(f) = (105.4 -100) X = 19.6 

So when we have the freezing point depression as 19.6,

we know water freezes at 0° C so the freezing point will be 0 - 19.6 = -19.6 °C

c (diamond),

sio2 (quartz)

a pan containing water is placed in a positio n where it is exposed to a 10 mile per hour wind. the pan contains a uniform depth of 1.27 cm of water, is quite wide, and has a length of 2.44 m in the wind direction. the water in the pan is at a constant temperature of 15.5 °c.