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Home » If 1.202 g mL^{-1}mL−1 is the density of 20% aqueous KI, determine the following:
If 1.202 g mL^{-1}mL−1 is the density of 20% aqueous KI, determine the following:
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If 1.202 g mL^{-1}mL−1 is the density of 20% aqueous KI, determine the following:

(a) Molality of KI

(b) Molarity of KI

(c) Mole fraction of KI

(a) Molar mass of \mathrm{Kl}=39+127=166 \mathrm{~g} \mathrm{~mol}^{-1}

20 \% aqueous solution of KI shows that 20 \mathrm{~g} of \mathrm{Kl} is present in 100 \mathrm{~g} of solution.

i.e.,

20 \mathrm{~g} of \mathrm{Kl} is present in (100-20) \mathrm{g} of water =80 \mathrm{~g} of water

Hence, molality of the solution =\frac{\text { Moles of KI }}{\text { Mass of water in } \mathrm{kg}}

=\frac{\frac{20}{166}}{0.08}m

=1.506~\text{m}

=1.51~\text{m (approx}\text{.) }

(b) The destiny of the solution is given that =1.202 g m L^{-1}

Volume of 100 \mathrm{~g} solution =\frac{\text { Mass }}{\text { Density }}

=\frac{100~\text{g}}{1.202~\text{g}~\text{m}{{\text{L}}^{-1}}}

=83.19~\text{mL}

=83.19 \times 10^{-3} L

Hence, molarity of the solution =\frac{\frac{20}{165} \mathrm{~mol}}{83.19 \times 10^{-3} L}

=1.45 \mathrm{M}

(c) Moles of \mathrm{KI}=\frac{20}{166}=0.12 \mathrm{~mol}

Moles of water =\frac{80}{18}=4.44 \mathrm{~mol}

Hence, mole =\frac{\text { Moles of } K I}{\text { Moles of } K I+\text { Moles of water }}

Fraction of \mathrm{KI}=\frac{0.12}{0.12+4.44} =0.0263

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