Calculate(a)molality (b)molarity and (c)mole fraction of KI if the density of 20% (mass/mass) aqueous KI is1.202 g mL^-1.

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Solution

(a)Molar mass of KI = 39 + 127 = 166 g mol⁻¹
20%(mass/mass) aqueous solution of KI means 20g of KI is present in 100 g of solution.
That is,
20g of KI is present in (100 − 20) g of water = 80 g of water
Therefore, molality of the solution $= \frac{m o l e s o f K I}{M a s s o f w a t e r i n K g}$
$= \frac{\frac{20}{166}}{0.08}$
$= 1.506 m$
$= 1.51 m$ (approximately)
(b) It is given that the density of the solution = 1.202 g mL⁻¹
∴Volume of 100 g solution $= \frac{m a s s}{D e n s i t y} = \frac{100}{1.202} = 83.19 m l$
=83.19 × 10⁻³L
Therefore, molarity of the solution $= \frac{\frac{20}{166} m o l e}{83.19 \times 10^{- 3} L}$
= 1.45 M
(c)Moles of KI $= \frac{20}{166} = 0.12 m o l$
Moles of water $= \frac{80}{18} = 4.44 m o l$
Therefore, mole fraction of KI $= \frac{m o l e s o f K I}{m o l e s o f K I + m o l e s o f w a t e r}$
$= \frac{0.12}{0.12 + 4.44} = 0.0263$

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Calculate(a)molality (b)molarity and (c)mole fraction of KI if the density of 20% (mass/mass) aqueous KI is1.202 g mL-1.

Calculate(a)molality (b)molarity and (c)mole fraction of KI if the density of 20% (mass/mass) aqueous KI is1.202 g mL^-1.