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A solution of \mathrm{2.5\ g} of a non-volatile solute in \mathrm{50\ g} of water was found to boil at \mathrm{101.4\ ^{\circ}C}. What is the elevation in boiling point of the solution? (Molar mass of water = \mathrm{18\ g/mol}, Kb of water = \mathrm{0.52\ ^{\circ}C/m} )

Option: 1

\mathrm{0.46\ ^{\circ}C}


Option: 2

\mathrm{0.63\ ^{\circ}C}


Option: 3

\mathrm{0.98\ ^{\circ}C}


Option: 4

\mathrm{1.27\ ^{\circ}C}


Answers (1)

best_answer

The elevation in boiling point \Delta T_b can be calculated using the formula:


\Delta T_b = i \times K_b \times m

where i is the van't Hoff factor, K_b is the molal boiling point elevation constant of the solvent, and m is the molality of the solution.

Since the solute is non-volatile, its van't Hoff factor is 1. The molality of the solution can be calculated as follows:

Moles\ of \ solute = \frac{2.5\ g}{M_{solute}}


Moles\ of\ solvent = \frac{50\ g}{18\ g/mol} = 2.78\ mol

 

Molality\ (m) = \frac{\text{moles of solute}}{\text{mass of solvent (in kg)}} = \frac{\frac{2.5\ g}{M_{solute}}}{2.78\ kg} = \frac{0.142}{M_{solute}}

Substituting the values in the formula, we get:

\Delta T_b = 1 \times 0.52\ \mathrm{^{\circ}C/m} \times \frac{0.142}{M_{solute}} = \frac{0.0738}{M_{solute}}\ \mathrm{^{\circ}C}

We can now solve for M_{solute} using the boiling point elevation observed:

 

{M_{solute}}\ = \frac{0.0738}{\mathrm{1.4\ ^{\circ}C}} = 0.0527\ \mathrm{kg/mol}

 

Finally, we can calculate the elevation in boiling point:

\mathrm{101.4\ ^{\circ}C} - \mathrm{100\ ^{\circ}C} = \frac{0.0738}{M_{solute}}\ \mathrm{^{\circ}C}
 

Therefore, the correct option isc) \mathrm{0.98\ ^{\circ}C}.

Note: The molar mass of the solute is not required to solve this problem.

 

Posted by

Pankaj Sanodiya

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