Henna is investigating the melting point of different salt solutions. She makes a salt solution using 10 mL of water with a known mass of NaCl salt. She puts the salt solution into a freezer and leaves it to freeze. She takes the frozen salt solution out of the freezer and measures the temperature when the frozen salt solution melts. She repeats each experiment.
Assuming the melting point of pure water as 0^oC, answer the following questions:
(a) One temperature in the second set of results does not fit the pattern. Which temperature is that? Justify your answer.
(b) Why did Henna collect two sets of results?
(c) In place of NaCl, if Henna had used glucose, what would have been the melting point of the solution with 0.6 g glucose in it?
OR
What is the predicted melting point if 1.2 g of salt is added to 10 mL of water? Justify your answer.

Question

Henna is investigating the melting point of different salt solutions. She makes a salt solution using 10 mL of water with a known mass of NaCl salt. She puts the salt solution into a freezer and leaves it to freeze. She takes the frozen salt solution out of the freezer and measures the temperature when the frozen salt solution melts. She repeats each experiment.
Assuming the melting point of pure water as 0^oC, answer the following questions:
(a) One temperature in the second set of results does not fit the pattern. Which temperature is that? Justify your answer.
(b) Why did Henna collect two sets of results?
(c) In place of NaCl, if Henna had used glucose, what would have been the melting point of the solution with 0.6 g glucose in it?
OR
What is the predicted melting point if 1.2 g of salt is added to 10 mL of water? Justify your answer.

Infinity Learn · Accepted Answer

(a) The melting point of ice is the same as the freezing point of water.The concept of depression in the freezing point property is used here. When the amount of salt is increased there is an increase in the depression of the freezing point and a decrease in melting point but the trend is not followed in this case 3rd reading for 0.5 g.(b) The average value of two sets of reading avoids errors in data.(c) ΔTf (glucose) = 1 x Kf x 0.6 x 1000/ 180 x 10ΔTf (NaCl) = 2 x Kf x 0.6 x 1000/ 58.5 x 103.8 = 2 x Kf x 0.6 x 1000 /58.5 x 10Divide equation 1 by 2ΔT (glucose0=/3.8 = 58.5/2 x 180ΔTf (glucose = 0.62 Freezing point or Melting point = – 0.62 oCORDepression in freezing point ∞ Molality0.3 g depression is 1.9 oC0.6 g depression is 3.8 oC1.2 g depression will be 3.8 x 2 = 7.6 oC

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