Question

Water has a specific heat of 4.184 J/g-degree. How many joules are needed to heat 525g of water from 25.0oC to 95.0oC ? Explaining why, which, if any, of the following water solutions will have the lowest freezing temperature ? Explain the reason for your answer.

A. 0.20m NaCl

B. 0.10m AlCl3

C. 0.40m KBr

D. 0.20m CaCl2

Answer 1

Amount of heat required,q =mcdt

Where m = mass of water=525 g

C = specific heat of water= 4.184J/goc

dt= change in temperature=95-25=70 oc

Plug the values we get q= 153.7x10^3 J

= 153.7kJ

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We know that ΔT f = iKf x m
Where
ΔT f = depression in freezing point
= freezing point of pure solvent – freezing point of solution
=
K f = depression in freezing constant
i= vanthoff’s factor =
m = molality of the solution

As Tf is more it has lowest freezing point

Which is proportional to ixm

A. 2x0.20=0.4

B. 4x0.1=0.4

C. 2x0.40=0.8

D. 3x0.20=0.6

So option C is correct