Calculate the change in the freezing point of an aqueous solution of 6.44 grams of glucose...
Calculate the change in the freezing point of an aqueous solution of 6.44 grams of glucose in 122 grams of water. The cryoscopic constant of water is 1.86 degrees centigrade / molality
Homework Answers
∆Tf = kf × molality
= 1.86°C/molal × (6.44g/180.156g/mol)×(1/0.122 kg)
= 1.86 ×0.293 °C
= 0.545 °C
The change in frezzing point of aqueous solution
= 0.545 °C ( Answer)
The freezing point of solution = - 0.545°C
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