What is the freezing point 1.56m acqueous solution of calcl2?
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Supposing complete ionization:
CaCl2 → Ca{2+} + 2 Cl{-} [three ions total]
(1.56 m CaCl2) x (3 mol ions / 1 mol CaCl2) = 4.68 m ions
(1.86 °C/m) x (4.68 m) = 8.70 °C change
0°C - 8.70°C = - 8.70°C
CaCl2 → Ca{2+} + 2 Cl{-} [three ions total]
(1.56 m CaCl2) x (3 mol ions / 1 mol CaCl2) = 4.68 m ions
(1.86 °C/m) x (4.68 m) = 8.70 °C change
0°C - 8.70°C = - 8.70°C
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