Question

If x is the mole fraction of a solute when dissolved in pure waiter, then prove that the molality of solution is

Molality (m) = (500x)/(9-9x)

PLEASE HELP ITS URGENT

Answer 1

i typed it into my fancy calculator an it said its 0 idk if its rite an idk how to do it

Answer 2

Hi, have been trying to work it out but am a little stuck. Here is what I have so far (need to go to bed!) but I think I'm on the right track, probably has a few errors but hopefully you can do the rest :)

m= n/v
x = n/m
x = number of sol molecules (n) / number of total molecules (n + w)
M = number of sol molecules (n) / volume (v)
M = (500x)/(9-9x)
n/v = 500x/(9-9x)
number of molecules in a volume of water = ?
room temp (water) 1g = 1ml
1L = 1000mL
water density = 1000g/L
1000g/L / 18g/mole = 55.555

using Avagadro
55.555 * (6.022*10^23) = 334.55 * 10^23 molecules in 1L of water

x = n/(n+ 3.34*10^25)
xn + 3.34*10^25x = n
3.34*10^25x= n - xn
3.34*10^25x = n(1-x)
n = 3.34*10^25x/(1-x)

M= n/v
M = (3.34*10^25x/(1-x))/V
now as we were using V = 1L
3.34*10^25x/(1-x) = (500x)/(9-9x)