Solve. An aqueous solution of calcium chloride (CaCl2) boils at 101.3 degrees Celsius. How many kilograms of calcium chloride were dissolved in 1000.0 g of the solvent?
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Im assuming you are familiar with the formula: ΔT_b = K_b * m * i. Where ΔT_b is hte change in boiling point, K_b is a constant determined by the solvent (For water K_b is 0.512 Celsius/m, and i is the number of particles the solute breaks into when it dissolves (for CaCl2 the i factor is 3 you know this becasue an an ionic substance, it breaks down into 1 Ca and 2 Cl, molecular subsances always have an i factor of 1).
You are trying to find the molality opf the subtance. (m= moles of solute/ kg of solvent).
1.3= 0.512 C/m * m * 3
1.3=1.536m
0.85=m
Now set up a porportion
x/1 kg = .85
x=.85 moles.
Molar mass of CaCl2 is 110.98 g per mole.
.85 * 110.98 g/mol = 94 grams.
94 grams of CaCl2 were dissolved in the solvent
You are trying to find the molality opf the subtance. (m= moles of solute/ kg of solvent).
1.3= 0.512 C/m * m * 3
1.3=1.536m
0.85=m
Now set up a porportion
x/1 kg = .85
x=.85 moles.
Molar mass of CaCl2 is 110.98 g per mole.
.85 * 110.98 g/mol = 94 grams.
94 grams of CaCl2 were dissolved in the solvent