.1 mol of c2h5oh
.1 mol of libr
.2 mol of c6h12o6
.2 mol of cacl2
1 Answer
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It will be the 0.2 mole of CaCl2, it ionizes and puts 3 ions x 0.2 mole (0.6 mole of ions) into the soulution. C2H5OH and C6H12O6 don’t ionize so only 0.1 & 0.2 mole of dissolved material for lowering freezing point. LiBr does ionize, but only 0.1 mole x 2 ions available.