Question

How many grams of BaCl2 should be added to 300g H2O so that the freezing point of the solution is lowered to -8.3 degrees celsius? Assume that the BaCl2 completely dissociates in the solution. Thanks!

Answer 1

92,8 g

Explanation:

`Delta T = i xx k_c xx m`
if `BaCl_2` is completely dissociated i=3; k for water is 1,86 °C/mol m is the molality , i.e. the number of moli of the salt in 1000 g of water
`m= (DeltaT)/ (Kc xx i)= (8.3°C) / (3 xx 1,86 °C/(mol))=1,48` mol of `BaCl_2` in 1000 g of water
hence 0.446 mol of salt in 300 g of water
Since MM `BaCl_2` is 208g/mol you need 92,8 g
the croscopy law is true for diluited concentration and this solution isn't and you must use a experimental nomogram to solve your problem if you want avoid big mistakes