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

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Solution

Molality
$\frac{w}{w} % = 20$
$100 g m$ solution has $20 g m$ KI
$80 g m$ solvent has $20 g m$ KI
$m = \frac{\frac{20}{166}}{\frac{80}{1000}} = 1.51 m o l$
Molarity
Volume = $\frac{m}{d} \Rightarrow \frac{20}{1.2} = 83.2$
Molalrity $= \frac{.120 \times 1000}{83.2}$
$= 1.44 m$
Mole fraction $= \frac{\frac{20}{166}}{\frac{80}{18} + \frac{20}{166}}$

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Calculate (a) molality (b) molarity and(c) mole fraction of KI if the density of 20% (mass/ mass) aqueous KI is 1.202 g mL−1

Molality ww%=20 100 gm solution has 20 gm KI 80 gm solvent has 20 gm KI m=20166801000=1.51mol Molarity Volume = md⇒201.2=83.2 Molalrity =.120×100083.2 =1.44 m Mole fraction =201668018+20166

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