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प्रश्न
Why is boiling point of 1 M NaCl solution more than that of 1 M glucose solution?
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उत्तर
Because the van't Hoff factor of 1 M NaCl (2) solution is bigger than that of 1 M glucose (1) solution, boiling point is directly proportional to molality and van't Hoff factor. As a result, the boiling point of 1 M NaCl is higher than that of 1 M glucose solution.
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संबंधित प्रश्न
The substance ‘X’, when dissolved in solvent water gave molar mass corresponding to the molecular formula ‘X3’. The van’t Hoff factor (i) is _______.
(A) 3
(B) 0.33
(C) 1.3
(D) 1
3.9 g of benzoic acid dissolved in 49 g of benzene shows a depression in freezing point of 1.62 K. Calculate the van't Hoff factor and predict the nature of solute (associated or dissociated).
(Given : Molar mass of benzoic acid = 122 g mol−1, Kf for benzene = 4.9 K kg mol−1)
How will you convert the following in not more than two steps:
Benzoic acid to Benzaldehyde
Give reasons for the following
Elevation of the boiling point of 1 M KCl solution is nearly double than that of 1 M sugar solution.
The freezing point depression constant for water is 1.86° K Kg mol-1. If 5 g Na2SO4 is dissolved in 45 g water, the depression in freezing point is 3.64°C. The Vant Hoff factor for Na2SO4 is ______.
The van’t Hoff factor (i) accounts for ____________.
Van’t Hoff factor i is given by the expression:
(i) i = `"Normal molar mass"/"Abnormal molar mass"`
(ii) i = `"Abnormal molar mass"/"Normal molar mass"`
(iii) i = `"Observed colligative property"/"Calculated colligative property"`
(iv) i = `"Calculated colligative property"/"Observed colligative property"`
Van't Hoff factor I is given by expression.
What is the expected each water van't Hoff factor for and K4[F4(CN6)] when it completely dissociated in waters.
When 9.45 g of ClCH2COOH is added to 500 mL of water, its freezing point drops by 0.5°C. The dissociation constant of ClCH2COOH is x × 10−3. The value of x is ______. (Rounded-off to the nearest integer)
[\[\ce{K_{f(H_2O)}}\] = 1.86 K kg mol−1]
