Calculate the freezing point and the boiling point of a solution that contains 35 g of C12H22O11 in 370 g of water.
Moles of "C_{12}H_{22}O_{11}=\\frac{35}{342}=0.102moles"
Mass of water "= 370g = 0.37kg"
Molarity "= \\frac{0.102}{0.37}=0.28M\/kg"
Kf= 1.86°c
Depression = "0.28\u00d71.86=0.52\u00b0c"
Freezing point "=0-0.52=-0.52 \u00b0c"
Kb=0.52°c
Elevation = "0.28\u00d70.52=0.15"
Boiling point ="100+0.15=100.15\u00b0c"
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