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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
Define van’t Hoff factor.
19.5 g of CH2FCOOH is dissolved in 500 g of water. The depression in the freezing point of water observed is 1.0°C. Calculate the van’t Hoff factor and dissociation constant of fluoroacetic acid.
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.
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[Given: R = 0.082 L atm K–1 mol–1]
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]
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(Kf = 1.86 K Kg mol−1)
Consider the reaction
\[\begin{bmatrix}\begin{array}{cc}
\phantom{.......}\ce{CH3}\\
\phantom{....}|\\
\ce{CH3CH2CH2 - \overset{⊕}{N} - CH2CH3}\\
\phantom{....}|\\
\phantom{.......}\ce{CH3}
\end{array}\end{bmatrix}\]\[\ce{OH^- ->[Heat] ?}\]
Which of the following is formed in a major amount?
Calculate Van't Hoff factor for an aqueous solution of K3 [Fe(CN)6] if the degree of dissociation (α) is 0.852. What will be boiling point of this solution if its concentration is 1 molal? (Kb = 0.52 K kg/mol)
