How to calculate freezing point of solution?
DeltaT(f)=K(f) * m
From book K(f) = 5.12 C/m constant
and if m = mols solute/Kg solvent ----
how do I get there? Or am I'm working it incorrectly?
We're not given Benzene as C6H6 and as first year chem students anything "extra" like that is usually given in the question so it doesn't seem like mols are required.
Oh I don't know! These last two chapters have slaughter me and our final is in three days. Please help and explain.
Answer:
You are working it correctly. The reason they gave you no m.w. or formula for benzene is that benzene is the solvent. Molal freezing point depression is the phenomenon, so you are correct to work with mols solute per Kg of solvent.
You have 58.0g benzene, which is 0.0580 Kg solvent.
There are several ways to calculate mols of the gaseous solute.
Here, I prefer to let 4.00L given gas/(22.4L any gas at STP/mole) = (4.00L/22.4L/mol) @ STP
The gas is not at 273 K and 760mmHg (STP)
So, just re-proportion the temp and pressure:
(4.00L/22.4L/mol)*(273K/300K)*... = 0.15993 mol @ STP
I'll leave the rest to you.
*Be sure and check my math!*
pV = n RT
n =Moles of gas =pV / RT
T = 27 + 273 = 300 K
p = 748 / 760 = 0.984 atm
n = 0.984 x 4.00 / 0.0821 x 300 = 0.16 ( mole solute )
solvent is benzene
Mass of benzene = 58.0 g = 0.058 g
molality = 0.16 / 0.058 = 2.76
Delta T = 5.12 x 2.76 = 14.1 °C
Melting point of benzene = 5.5 °C
The freezing point of this solution will be
5.5 - 14.1 = - 8.62 °C
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