Saved Bookmarks
| 1. |
Calculate (a) molaity (b) molarity and (c) mole fraction of KI if the density of 20% (mass/mass) aqueous KI s "1.202 g mL"^(-1). |
|
Answer» Solution :`20%` (MASS/mass) aqueous KI solution means that Mass of KI = 20 g, `""` Mass of solution in water = 100 g `therefore"Mass of solvent (water) "=100-20=80g=0.080 KG` (a) Calculation of molality `"Molar mass of KI"=39+127="166 g mol"^(-1)"" therefore"Moles of KI"=("20 g")/("166 g mol"^(-1))=0.120` `"Molality of solution"=("No. of moles of KI")/("Mass of solvent in kg")=("0.120 MOLE")/("0.080 kg")="1.5 mol kg"^(-1)`. (b) Calculation of molarity `"Density of solution "="1.202 g mL"^(-1)""therefore"VOLUME of solution "=(100g)/("1.202 g mL"^(-1))="83.2 mL = 0.0832 L"` `"Molarity of solution"=("Moles of solute")/("Volume of solution in L")=("0.120 mole")/("0.0832 L")="1.44 M".` (c) Calculation of mole fraction of KI No. of moles of KI = 0.120 `"No. of moles of water"=("Mass of water")/("Molar mass of water")=("80 g")/("18 g mol"^(-1))=4.44` `"Mole fraction of KI"=("No. of moles of KI")/("Total no. of moles in solution")=(0.120)/(0.120+4.44)=(0.120)/(4.560)=0.0262`. |
|