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)`.

Answer

(a) 1.5 mol `kg^(-1)` (b) 1.44 mol `L^(-1)` (c )0.0263

Explanation

Mass of KI in 100 g solution = 20 g
Mass of water `= (100-20) g = 80g = 8.0 xx 10^(-2)` kg
Molar Mass of `KI = 39+ 127 = 166" g mol"^(-1)`
Molar Mass of water `= 18" g mol"^(-1)`
Number of moles of `KI=(80)/(166)=0.12` mol
Number of moles of water `=(80)/(18)=4.44` mol
Total number of moles `=0.12 +4.44= 4.56` mol
Volume of 100 g KI solution
`=(M)/(d)=(100g)/(1.202"g mL"^(-1))=83.2mL=83.2 xx 10^(-3)L`
(a) Molality `=("Moles of KI")/("Mass of water in kg")=(0.12"mol")/(8xx10^(-2)kg)`
`=1.5"mol kg"^(-1)`
(b) Molarity `=("Moles of KI")/("Volume of sol in litre")=(0.12"mol")/(83.2xx10^(-3)L)`
`=1.44"mol L"^(-1)`
(c ) Mole fraction of KI `=("No. of Moles of KI")/("Total no. of moles")=(0.12)/(4.56)`
= 0.0263