Question

A 5 percent of solution in water has a freezing point of 271k ,find the freezing of5 percent glucose .In water the freezing point in pure is273.15k intext:zigya

Answer 1

We know that :

ΔT = K * molality but we know that

molality is also defined in the form weight by weight ration into 10 by molecular weight of solute.

So, 2.15 = v$\frac{5*10}{weight of solute}$

Similarly, use the same thing for glucose , where weight of glucose is known (as glucose is solute).So, now you get ΔT, from that find freezing point of solution required (i.e. 5% glucose).

This was a simple question, asking this sort questions looks wierd as myself I am a Ph.D Scholar.

Thanking You.

Answer 2

Explanation:

Sample 'B' will not freeze at 0°C because it is not pure water. At one atmospheric pressure, the boiling point of pure water is 100°C and the freezing point of pure water is 0°C.