Calculate
(a) molality
(b) molarity and
(c) mole fraction of $K I$ if the density of $20 %$ (mass/ mass) aqueous $K I$ is $1.202 g m L^{- 1}$

Question

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Answer 1

Molality
$\frac{w}{w} % = 20$

$100 g m$ solution has $20 g m$ KI

$80 g m$ solvent has $20 g m$ KI

$m = \frac{\frac{2 0}{1 6 6}}{\frac{8 0}{1 0 0 0}} = 1.51 m o l$

Molarity
Volume = $\frac{m}{d} \Rightarrow \frac{2 0}{1 . 2} = 83.2$

Molalrity $= \frac{. 1 2 0 \times 1 0 0 0}{8 3 . 2}$

$= 1.44 m$

Mole fraction $$ = \dfrac{\dfrac{20}{166}}{\dfrac{80}{18} + \dfrac{20}{166}} $

Answer 2

Molality

\frac{w}{w} % = 20

100 g m

solution has

20 g m

KI

80 g m

solvent has

20 g m

KI

m = \frac{\frac{2 0}{1 6 6}}{\frac{8 0}{1 0 0 0}} = 1.51 m o l

Molarity
Volume =

\frac{m}{d} \Rightarrow \frac{2 0}{1 . 2} = 83.2

Molalrity

= \frac{. 1 2 0 \times 1 0 0 0}{8 3 . 2}

= 1.44 m

Mole fraction $$ = \dfrac{\dfrac{20}{166}}{\dfrac{80}{18} + \dfrac{20}{166}} $