2.5   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}.

Answers (1)

If we assume our solution is 100 g. Then according to question,  20 g KI is present and 80 g is water.

So moles of KI   :

                                             =\frac{20}{166}           \left ( Molar\ mass = 39+127 = 166\ g\ mol^{-1} \right )

(a) Molality :-  

                                 Molality = \frac{Moles}{Mass\ of\ solvent\ in\ Kg} = \frac{\frac{20}{166}}{0.08}= 1.506\ m.

(b) Molarity :-  

                                         Density = \frac{Mass}{Volume}

                                       Volume = \frac{Mass}{Density} = \frac{100}{1.202} = 83.19 mL                  

                                       Molarity = \frac{Moles}{Volume(l)} = \frac{\frac{20}{166}}{83.19\times10^{-3}} = 1.45\ M

 

(c)  Mol fraction :-   Moles of water :-

                                                                        = \frac{80}{18} = 4.44

                                  So, mol fraction of KI :-

                                                                            = \frac{0.12}{0.12+4.44} = 0.0263

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