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Assuming complete
Assuming complete
dissociation of the solute, how many grams of KNO3 must be added to 275 mL of
water to produce a solution that freezes at -14.5 C?
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Question 2 :
Assuming complete dissociation of the solute, how many grams of KNO3 must be added to 275 mL of water to produce a solution that freezes at -14.5 C?
The freezing point for pure water is 0.0 C and Kf is equal to 1.86 C/m.
Answer:
delta T = i x Kf x molality (i = 2 for KNO3 (KNO3 dissociates into 2 ions, K+ and [NO3]-). 14.5 = 2 x 1.86 x m molality = 3.90 = moles KNO3 / 0.275 kg H2O moles KNO3 = 3.90 x 0.275 = 1.07 moles KNO3 1.07 moles KNO3 x (101.1 g KNO3 / 1 mole KNO3) = 108 g KNO3
Question 2 :
Assuming complete dissociation of the solute, how many grams of KNO3 must be added to 275 mL of water to produce a solution that freezes at -14.5 C?
The freezing point for pure water is 0.0 C and Kf is equal to 1.86 C/m.
Answer:
delta T = i x Kf x molality (i = 2 for KNO3 (KNO3 dissociates into 2 ions, K+ and [NO3]-). 14.5 = 2 x 1.86 x m molality = 3.90 = moles KNO3 / 0.275 kg H2O moles KNO3 = 3.90 x 0.275 = 1.07 moles KNO3 1.07 moles KNO3 x (101.1 g KNO3 / 1 mole KNO3) = 108 g KNO3
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