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Mixtures and their Separation - How is it different from simple distillation?-advanced

Grade 9CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Simple Distillation is used for the separation of components of a mixture containing two miscible liquids that boil without decomposition and have a sufficient difference in their boiling points, typically ΔT>25 K\Delta T > 25\text{ K}.

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Fractional Distillation is employed when the difference in boiling points of the miscible liquids is less than 25 K25\text{ K}.

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The primary structural difference is the use of a fractionating column in fractional distillation, which is fitted between the distillation flask and the condenser.

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A fractionating column is usually a tube packed with glass beads. The beads provide a large surface area for the vapors to cool and condense repeatedly, allowing for better separation of liquids with close boiling points.

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In fractional distillation, the component with the lower boiling point distills over first, while the vapors of the higher boiling point component condense in the column and trickle back into the flask.

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Applications of Fractional Distillation include the separation of different fractions from petroleum products and the separation of various gases from liquid air (N2N_{2}, ArAr, O2O_{2}).

📐Formulae

ΔT=∣BP1−BP2∣\Delta T = |BP_{1} - BP_{2}|

If ΔT>25 K⇒Simple Distillation\text{If } \Delta T > 25\text{ K} \Rightarrow \text{Simple Distillation}

If ΔT<25 K⇒Fractional Distillation\text{If } \Delta T < 25\text{ K} \Rightarrow \text{Fractional Distillation}

T(K)=T(∘C)+273.15T(\text{K}) = T(^{\circ}\text{C}) + 273.15

💡Examples

Problem 1:

A mixture contains Ethanol (Boiling Point =78∘C= 78^{\circ}\text{C}) and Water (Boiling Point =100∘C= 100^{\circ}\text{C}). Determine which distillation method is suitable and calculate the difference in their boiling points in Kelvin.

Solution:

  1. Calculate the difference in Celsius: ΔT∘C=100∘C−78∘C=22∘C\Delta T_{^{\circ}\text{C}} = 100^{\circ}\text{C} - 78^{\circ}\text{C} = 22^{\circ}\text{C}
  2. Since a temperature interval of 1∘C1^{\circ}\text{C} is equal to an interval of 1 K1\text{ K}, the difference in Kelvin is: ΔT=22 K\Delta T = 22\text{ K}
  3. Since 22 K<25 K22\text{ K} < 25\text{ K}, Fractional Distillation must be used.

Explanation:

Because the boiling point difference is less than the threshold of 25 K25\text{ K}, simple distillation would result in a mixture of vapors. A fractionating column is required to provide multiple condensation-evaporation cycles to effectively separate ethanol from water.

Problem 2:

Calculate the boiling points of Nitrogen (−196∘C-196^{\circ}\text{C}) and Oxygen (−183∘C-183^{\circ}\text{C}) in Kelvin and determine if simple distillation can separate them from liquid air.

Solution:

  1. Convert Nitrogen BP to Kelvin: TN2=−196+273=77 KT_{N_{2}} = -196 + 273 = 77\text{ K}
  2. Convert Oxygen BP to Kelvin: TO2=−183+273=90 KT_{O_{2}} = -183 + 273 = 90\text{ K}
  3. Calculate the difference: ΔT=90 K−77 K=13 K\Delta T = 90\text{ K} - 77\text{ K} = 13\text{ K}

Explanation:

Since the difference ΔT=13 K\Delta T = 13\text{ K} is significantly less than 25 K25\text{ K}, simple distillation is insufficient. Fractional distillation is necessary to separate these gases from liquid air.