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प्रश्न
Calculate the freezing point of solution when 1.9 g of MgCl2 (M = 95 g mol−1) was dissolved in 50 g of water, assuming MgCl2 undergoes complete ionization.
(Kf for water = 1.86 K kg mol−1)
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उत्तर
Given:
Kf = 1.86 K kg mol−1
Mass of solute = 1.9 g
Mass of solvent = 50 g
Therefore,
`"Molality of the solution, "m=1.9/95xx1000/50=0.4m`
Also, MgCl2 undergoes complete ionisation and thereby yielding 3 moles of constituent ions for every mole of MgCl2.
∴ i = 3
Now, depression in freezing point is given as
∆Tf = iKfm
= 3×1.86×0.4
= 2.232 K
Tf = 273.15−2.232
=270.918 K
Hence, the new freezing point of the solution is 270.92 K.
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Which of the following statements is false?
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 |
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- Why did Henna collect two sets of results?
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OR
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