The freezing point of the 0.05 molal solution of non - electrolyte in water is? (kf = 1.86)
Solution and Explanation
ΔTf = Kf x m
Given: Kf = 1.86 (in units of molal per degree Celsius)
m = 0.05 molal
Substituting the given values into the formula, we get:
ΔTf = 1.86 x 0.05
ΔTf = 0.093oC
The freezing point depression (ΔTf) is 0.093oC
Freezing point of the solution = Freezing point of water - ΔTf
Freezing point of the solution = 0oC - 0.093oC = -0.093oC
Freezing point of the 0.05 molal solution of the non-electrolyte in water is approximately -0.093oC.
freezing point (ΔTf) can be calculated using the formula ΔTf=kf×m, where kf is the cryoscopic constant and m is the molality. For the given values kf =1.86 and m=0.05, the calculation yields ΔTf=0.093.
Thus, the final freezing point (Tf) is found by subtracting 0.0930.093 from the initial temperature,
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Concepts Used:
Concentration of Solutions
It is the amount of solute present in one liter of solution.
Concentration in Parts Per Million - The parts of a component per million parts (106) of the solution.
Mass Percentage - When the concentration is expressed as the percent of one component in the solution by mass it is called mass percentage (w/w).
Volume Percentage - Sometimes we express the concentration as a percent of one component in the solution by volume, it is then called as volume percentage
Mass by Volume Percentage - It is defined as the mass of a solute dissolved per 100mL of the solution.
Molarity - One of the most commonly used methods for expressing the concentrations is molarity. It is the number of moles of solute dissolved in one litre of a solution.
Molality - Molality represents the concentration regarding moles of solute and the mass of solvent.
Normality - It is the number of gram equivalents of solute present in one liter of the solution and it is denoted by N.
Formality - It is the number of gram formula present in one litre of solution.
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