1.

How many grams of KCI should be added to 1.0 kg of water to lower its freezing point to -8^(@)C. Calculate the degree of molality of BaCI_(2) in this solution. (K_(b) "for water "=0.52 K m^(-1))

Answer»


Solution :In solution, KCI dissociates as :
`KCIoverset(("AQ"))toK^(+)CI^(-)("aq")therefore" Van't Hoff factpr"(i)=2`
`DeltaT_(f)=0^(@)C-(-08^(@)C)=8^(@)C=8C=8K, *K, K_(f)=1.86" K KG mole"^(-1)`
`DeltaT_(f)=ixxK_(f)xxm or m=(DeltaT_(f))/(ixxK_(f))`
`m=((8K))/(2xx(1.86"K kg mol"^(-1)))=2.15" mol kg"^(-1)=2.15m`
`therefore"Mass of KCI in grams = (2.15 mol)XX(74.5" g mole "^(-1))=162.2 g.`


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