1.

0.01 M solution of KCl and BaCl_(2) are prepared in water. The freezing points of KCl is found to be -2^(@)C. What is the freezing point of BaCl_(2) to be completely ionized

Answer»

`-3^(@)C`
`+3^(@)C`
`-2^(@)C`
`-4^(@)C`

Solution :i for KCl = 2, i for `BaCl_(2)=3 because Delta T_(F) PROP i`
`(Delta T_(f)(KCl))/(Delta T_(f)(BaCl_(2)))=(2)/(3)rArr Delta T_(f)(BaCl_(2))=(3)/(2)xx2=3^(@)C`
`therefore` Freezing point of `BaCl_(2)=-3^(@)C`.


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