Question

1)0.5m Li2SO4 2)0.5 KCL 3)0.5 Al2(SO4)3 4)0.1m Bacl2 find out highest boiling point,highest freezing point,lowest vapour pressure, lowest π​

Answer 1

Answer:

HELLO FRIEND HERE IS YOUR ANSWER,,,,,,

Freezing point is the temperature at which the liquid state and the solid state of the same substances are in a perfect equilibrium and should,, hence,, have the same vapour pressure. Due to the attribution of lower vapour pressure of the solution quantity, the solid from a solution separates out at a even lower temperature as shown in my attachment above (in app) or below (in website).

Note: Excuse my Lackadaisically written handwriting.

Here, the answer should be 0.5 M or 0.5 Molar

Al\supersuperscript{2}\supersuperscript2 (SO\supersuperscript{4}\supersuperscript4 )\supersuperscript{3}\supersuperscript3 \implies⟹ 2Al \supersuperscript{3}\supersuperscript3 (+) + 3(SO\supersuperscript{4}\supersuperscript4 )\supersuperscript{2}\supersuperscript2

This colligative property will be directly proportional to the Van't Hoff's factor of a applied osmotic pressure and depression in freezing point. The more and higher the value of Van't Hoff factor goes above with a same concentration the colligative property will also increase by it for the same.

In this particular case the Van't Hoff factor is showing the advent of 5 discretion of ionic compounds by addition of 2 + 3 ions supporting them in a balanced equation. So therefore the Van't Hoff factor in Al\supersuperscript{2}\supersuperscript2 (SO\supersuperscript{4}\supersuperscript4 )\supersuperscript{3}\supersuperscript3 is \bold{5}5 . All five ions are produced inside a solution.

Essentially the other compounds have Van't Hoff factors of;;

KCl is K (+) and Cl (-) which are undergoing ionisation and giving out two ions or approximately 1.86 ions.

Van't Hoff factor for BaCl\supersuperscript{2}\supersuperscript2 ,

BaCl\supersuperscript{2}\supersuperscript2 \implies⟹

Ba (2+) + 2Cl (-), the Van't Hoff factor will be "3" since it has been dissociated or the discretion of 3 ions has been done.

For Li\supersuperscript{2}\supersuperscript2 SO\supersuperscript{4}\supersuperscript4 again the Van't Hoff factor will be "2" as Lithium has 2 ions on its side but for Sulphate complex, it has no ions or we can say one ion that will be "3" or between "2" and "3".

HOPE THIS HELPS AND GIVES YOU THE DISTINCTIONS OF VAN'T HOFF FACTOR USES IN FREEZING POINTS!!!!!