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If a hot, saturated solution of copper sulfate is cooled rapidly in ice-cold water, smaller and less well-formed crystals will form than if it is cooled slowly at room temperature.

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प्रश्न

If a hot, saturated solution of copper sulfate is cooled rapidly in ice-cold water, smaller and less well-formed crystals will form than if it is cooled slowly at room temperature. How would you design and perform an experiment to test this hypothesis?

Hint: Prepare a hot saturated solution of copper sulfate and divide it into two equal parts.

विस्तार में उत्तर
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उत्तर

Aim: To investigate how the rate of cooling influences the size and shape of copper sulfate crystals.

Procedure:

  1. A hot, saturated solution of copper sulfate is prepared by dissolving it in hot water, and a few drops of dilute sulfuric acid are added to obtain pure crystals.
  2. The solution is then filtered to remove any insoluble impurities. It is divided equally into two beakers, A and B.
  3. Beaker A is cooled quickly by placing it in ice-cold water, while beaker B is left to cool slowly at room temperature without disturbance.
  4. After some time, the crystals formed in both beakers are observed and compared.

Observation: Rapid cooling in beaker A produces small, irregular crystals, whereas slow cooling in beaker B results in larger, shiny, and well-formed crystals.

Conclusion: Slow cooling allows particles sufficient time to arrange themselves in an orderly pattern, resulting in the formation of larger and well-shaped crystals. This supports the hypothesis that rapid cooling leads to smaller and less well-formed crystals compared to slow cooling.

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अध्याय 5: Exploring Mixtures and their Separation - Intext Questions [पृष्ठ ७९]

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एनसीईआरटी Science Exploration [English] Class 9
अध्याय 5 Exploring Mixtures and their Separation
Intext Questions | Q 1. | पृष्ठ ७९
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