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प्रश्न
Calculate the freezing point of the solution when 31 g of ethylene glycol (C2H6O2) is dissolved in 500 g of water.
(Kf for water = 1.86 K kg mol–1)
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उत्तर
The freezing point of pure water is 273.15 K. On dissolving ethylene glycol, freezing point, being a colligative property, will be lowered.
`DeltaT_f=K_"f"(w_sxx1000)/(M_sxxW`
We are given that
Kf for water=1.86 K kg mol−1
Mass of solute, ws=31 g
Molar mass of solute, Ms = 12 × 2 + 1 × 6 + 16 × 2
= 62 g mol−1
Mass of water, W=500 g
`thereforeDeltaT_f=1.86xx(31xx1000)/(62xx500)`
= 1.86 K
Hence, the freezing point of the solution,
Tf = 273.15 K−1.86 K
= 271.29 K
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| 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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