Question

Calculate molality,molarity and mole fraction of KI if the density of 20% w/w aqueous KI is 1.202g/ml.

Answer 1

Density of solution 1.202 g/ml

concentration of KI = 20% weight by weight

This means that:-

There is 20g of Potassium Iodide present in 100g of the aqueous solution.

now:-

volume of solution will be :- mass divided by density

hence:-

$volume=\dfrac{100}{1.202}\\ \\ \\ \\V=83.19ml$

now:-

moleular mass of KI = 39+127=166 u

Number of moles of KI = 20/166=0.12

molecular mass of water= 2+16=18

number of moles of water= 100/18=5.55

now:-

MOLE FRACTION :-

mole fraction of KI is equal to :-

$=\dfrac{0.12}{0.67}\\ \\ \\=0.18$

MOLARITY :-

molarity of KI will be :-

$=\dfrac{20g}{0.83L}\\ \\ \\=24.1\:gram/litre$

MOLALITY :-

Molality of KI will be :-

$=\dfrac{0.12moles}{0.10kg}\\ \\ \\=1.2\:moles/kilogram$

NOTE :-

since the numbers do not cancel each other ,the concept of rounding off significant figures has been used.

The answer may differ to nearest decimal place as the calculation is done by rounding off correct to two decimal places.

there will be minor differences only!

These differences are acceptable in the stoichiometric calculations in chemistry!