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For a dilute solution containing 2.5 g of a non-volatile non-electrolyte solute in 100 g of water, the elevation in boiling point at 1 atm pressure is 2°C. Assuming concentration of solute is much lower than the concentration of solvent, the vapour pressure (mm of Hg) of the solution is (take Kb = 0,76 K kg mol-1).
(2012)
- a)724
- b)740
- c)736
- d)718
Correct answer is option 'A'. Can you explain this answer?
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To calculate the elevation in boiling point, we can use the formula:
ΔTb = Kb * m
Where:
ΔTb = elevation in boiling point
Kb = molal boiling point elevation constant
m = molality of the solution
First, we need to calculate the molality of the solution.
Molality (m) is defined as the number of moles of solute per kilogram of solvent.
Given:
Mass of solute (m_solute) = 2.5 g
Mass of solvent (m_solvent) = 100 g
Molar mass of water (M_water) = 18 g/mol
To find the number of moles of solute (n_solute), we can use the formula:
n_solute = m_solute / M_solute
We are given that the solute is non-volatile and non-electrolyte, so its molar mass is equal to its formula mass.
Assuming the molar mass of the solute is M_solute, we can calculate the value of n_solute:
n_solute = 2.5 g / M_solute
To find the molality (m), we can use the formula:
m = n_solute / m_solvent
Substituting the values we have:
m = (2.5 g / M_solute) / 100 g
Now, we can substitute the value of molality (m) into the formula for the elevation in boiling point:
ΔTb = Kb * m
We are given that the elevation in boiling point is 2 °C (or 2 K). So, the equation becomes:
2 K = Kb * [(2.5 g / M_solute) / 100 g]
To solve for Kb, we need to rearrange the equation:
Kb = (2 K) / [(2.5 g / M_solute) / 100 g]
Simplifying further:
Kb = (200 g * M_solute) / 2.5 g
Kb = 80 M_solute
Therefore, the molal boiling point elevation constant (Kb) is 80 M_solute.
ΔTb = Kb * m
Where:
ΔTb = elevation in boiling point
Kb = molal boiling point elevation constant
m = molality of the solution
First, we need to calculate the molality of the solution.
Molality (m) is defined as the number of moles of solute per kilogram of solvent.
Given:
Mass of solute (m_solute) = 2.5 g
Mass of solvent (m_solvent) = 100 g
Molar mass of water (M_water) = 18 g/mol
To find the number of moles of solute (n_solute), we can use the formula:
n_solute = m_solute / M_solute
We are given that the solute is non-volatile and non-electrolyte, so its molar mass is equal to its formula mass.
Assuming the molar mass of the solute is M_solute, we can calculate the value of n_solute:
n_solute = 2.5 g / M_solute
To find the molality (m), we can use the formula:
m = n_solute / m_solvent
Substituting the values we have:
m = (2.5 g / M_solute) / 100 g
Now, we can substitute the value of molality (m) into the formula for the elevation in boiling point:
ΔTb = Kb * m
We are given that the elevation in boiling point is 2 °C (or 2 K). So, the equation becomes:
2 K = Kb * [(2.5 g / M_solute) / 100 g]
To solve for Kb, we need to rearrange the equation:
Kb = (2 K) / [(2.5 g / M_solute) / 100 g]
Simplifying further:
Kb = (200 g * M_solute) / 2.5 g
Kb = 80 M_solute
Therefore, the molal boiling point elevation constant (Kb) is 80 M_solute.
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For a dilute solution containing 2.5 g of a non-volatile non-electrolyte solute in 100 g of water, the elevation in boiling point at 1 atm pressure is 2°C. Assuming concentration of solute is much lower than the concentration of solvent, the vapour pressure (mm of Hg) of the solution is (take Kb = 0,76 K kg mol-1).(2012)a)724b)740c)736d)718Correct answer is option 'A'. Can you explain this answer?
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