The boiling point of benzene is 353 . 23 K . When 1 . 80 g of non-volatile solute was dissolved in 90 g of benzene, the boiling point of is raised to 354 . 11 K . Calculate the molar mass of the solute. ( K b for benzene is 2 . 53 K Kg mol - 1 )

Given, The boiling point of pure benzene = 353 . 23 K The boiling point of the solution = 354 . 11 K The weight of the benzene W A = 90 g The weight of the solute W B = 1 . 80 g Molal boiling point elevation constant K b = 2 . 53 K Kg mol - 1 So, the elevation in boiling point ∆ T b = 354 . 11 - 353 . 23 = 0 . 88 K If the molar mass of the solute is M B , then ⇒ M B = K b × W B × 1000 ∆ T b × W A ⇒ M B = 2 . 53 × 1 . 80 × 1000 0 . 88 × 90 = 57 . 5 g mol - 1 The molar mass of the solute is 57 . 5 g mol - 1 .

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All In One ISC Chemistry Class 12>Chapter 2 - Solutions>Chapter Practice>Q 54

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How does Van't Hoff factor helps in the determination of degree of association or dissociation of a solute in solution?

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All In One ISC Chemistry Class 12>Chapter 2 - Solutions>Chapter Practice>Q 55

When dehydrated fruits and vegetables are placed in water, they slowly swell and return to their original shape. Why? What is effect of temperature on this process? Explain.