The molal boiling point constant for water is 0.513⁰ C Kg/mol. When 0.1 mole of sugar is dissolved in 200mg of water, the solution boils under a pressure of one atmosphere at ___
Answer
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Hint: The temperature at which a liquid's vapour pressure equals the pressure of the gas above it is known as the boiling point. A substance's boiling point is the temperature at which the vapour pressure of a liquid equals the pressure surrounding the liquid and the liquid transforms into a vapour. The boiling point of a liquid is affected by the surrounding atmospheric pressure.
Complete answer:
The basic equation for calculating a solution's boiling point is \[\Delta T{\text{ }} = {\text{ }}m{K_b}\]. The boiling-point elevation, meaning how much higher the boiling point of the solution is than that of the pure solvent, is referred to as \[\Delta T{\text{ }}\]. The temperature is measured in degrees Celsius. The molal boiling-point elevation constant is known as \[{K_b}\]. When one mole of solute is dissolved in one kg of solution, the boiling point rises. The molal boiling point elevation constant, \[{K_b}\] is the proportionality constant.
Now from the question,
\[{K_b}\] = 0.513⁰ C Kg/mol
\[\Delta T{\text{ }}\]= 100⁰ C
Given, moles of sugar = 0.1
Mass of Water = 200 ml = 0.2 l
Molality is the number of moles of solute per kg of liquid which is a characteristic of a solution.
Using the formula Molality = mol/kg
Molality, \[m = \dfrac{{0.1}}{{0.2}}\]
m = 0.5
Now, substituting all the values,
\[\Delta T{\text{ }} = {\text{ }}m{K_b}\]
\[\Delta T{\text{ }}\]= \[{T_b} - {T_b}^0\]
\[{T_b} - {T_b}^0 = m{K_b}\]
\[{T_b}\] - 100 = 0.5 x 0.513
\[{T_b}\]= 100 + 0.256
\[{T_b}\]= 100.256⁰ C
Note:
The basic equation for calculating a solution's boiling point is \[\Delta T{\text{ }} = {\text{ }}m{K_b}\]. The boiling-point elevation, meaning how much higher the boiling point of the solution is than that of the pure solvent, is referred to as \[\Delta T{\text{ }}\]. The temperature is measured in degrees Celsius. The molal boiling-point elevation constant is known as \[{K_b}\]. When one mole of solute is dissolved in one kg of solution, the boiling point rises. The molal boiling point elevation constant, \[{K_b}\] is the proportionality constant.
Complete answer:
The basic equation for calculating a solution's boiling point is \[\Delta T{\text{ }} = {\text{ }}m{K_b}\]. The boiling-point elevation, meaning how much higher the boiling point of the solution is than that of the pure solvent, is referred to as \[\Delta T{\text{ }}\]. The temperature is measured in degrees Celsius. The molal boiling-point elevation constant is known as \[{K_b}\]. When one mole of solute is dissolved in one kg of solution, the boiling point rises. The molal boiling point elevation constant, \[{K_b}\] is the proportionality constant.
Now from the question,
\[{K_b}\] = 0.513⁰ C Kg/mol
\[\Delta T{\text{ }}\]= 100⁰ C
Given, moles of sugar = 0.1
Mass of Water = 200 ml = 0.2 l
Molality is the number of moles of solute per kg of liquid which is a characteristic of a solution.
Using the formula Molality = mol/kg
Molality, \[m = \dfrac{{0.1}}{{0.2}}\]
m = 0.5
Now, substituting all the values,
\[\Delta T{\text{ }} = {\text{ }}m{K_b}\]
\[\Delta T{\text{ }}\]= \[{T_b} - {T_b}^0\]
\[{T_b} - {T_b}^0 = m{K_b}\]
\[{T_b}\] - 100 = 0.5 x 0.513
\[{T_b}\]= 100 + 0.256
\[{T_b}\]= 100.256⁰ C
Note:
The basic equation for calculating a solution's boiling point is \[\Delta T{\text{ }} = {\text{ }}m{K_b}\]. The boiling-point elevation, meaning how much higher the boiling point of the solution is than that of the pure solvent, is referred to as \[\Delta T{\text{ }}\]. The temperature is measured in degrees Celsius. The molal boiling-point elevation constant is known as \[{K_b}\]. When one mole of solute is dissolved in one kg of solution, the boiling point rises. The molal boiling point elevation constant, \[{K_b}\] is the proportionality constant.
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