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प्रश्न
Read the passage carefully and answer the questions that follow:
| 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. |
| S.No | Mass of the salt used in g |
Melting point in °C | |
| Readings Set 1 | Reading Set 2 | ||
| 1 | 0.3 | -1.9 | -1.9 |
| 2 | 0.4 | -2.5 | -2.6 |
| 3 | 0.5 | -3.0 | -5.5 |
| 4 | 0.6 | -3.8 | -3.8 |
| 5 | 0.8 | -5.1 | -5.0 |
| 6 | 1.0 | -6.4 | -6.3 |
Assuming the melting point of pure water as 0°C, answer the following questions:
- One temperature in the second set of results does not fit the pattern. Which temperature is that? Justify your answer.
- Why did Henna collect two sets of results?
- 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.
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उत्तर
The melting point of ice is the freezing point of water. We can use the depression in freezing point property in this case.
a. 3rd reading for 0.5 g there has to be an increase in the depression of freezing point and therefore decrease in freezing point so also decrease in melting point when the amount of salt is increased but the trend is not followed in this case.
b. Two sets of reading help to avoid errors in data collection and give more objective data.
c. ΔTf (glucose) = `1 xx "K"_"f" xx (0.6 xx 1000)/(180 xx 10)`
ΔTf (NaCl) = `2 xx "K"_"f" xx (0.6 xx 1000)/(58.5 xx 10)`
3.8 = `2 xx "K"_"f" xx (0.6 xx 1000)/(58.5 xx 10)`
Divide equation 1 by 2
`(Δ"T"_"f" ("glucose"))/3.8 = 58.5/(2 xx 180)`
ΔTf (glucose) = 0.62
Freezing point or Melting point = - 0.62°C
OR
Depression in freezing point is directly proportional to molality (mass of solute when the amount of solvent remains same)
0.3 g depression is 1.9°C
0.6 g depression is 3.8°C
1.2 g depression will be 3.8 × 2 = 7.6°C
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