What is the freezing-point depression of water in a solution of 100g of sucrose, C12 H22 O11, and 500g of water? (Molar freezing point depression constant for water is -1.86°C/m.)
molality= molesSucrose/kgwater
= 100/(342*.5)=.58
deltaTf= m*(-1.86)=-1.09C
To find the freezing-point depression of water in a solution, you need to use the formula:
∆T = Kf * m
Where:
∆T is the freezing-point depression (in °C)
Kf is the molar freezing-point depression constant for water (-1.86 °C/m)
m is the molality of the solution (in mol/kg)
First, you need to determine the molality (m) of the sucrose solution. Molality is defined as the number of moles of solute per kilogram of solvent. In this case, the solute is sucrose (C12H22O11), and the solvent is water.
To find the moles of sucrose, you divide the given mass by the molar mass of sucrose:
Molar mass of sucrose (C12H22O11) = 12 * 12.01 (C) + 22 * 1.01 (H) + 11 * 16.00 (O) = 342.34 g/mol
Moles of sucrose = mass of sucrose / molar mass of sucrose
Moles of sucrose = 100 g / 342.34 g/mol
Next, you calculate the molality (m) using the moles of solute and the mass of the solvent:
Molality (m) = moles of solute / mass of solvent (in kg)
Mass of solvent = 500 g = 500 g / 1000 g/kg
Now that you have the molality (m), you can calculate the freezing-point depression (∆T) using the formula mentioned earlier:
∆T = Kf * m
Substituting the values:
∆T = -1.86 °C/m * molality (m)
By calculating the values as explained above, you can find the freezing-point depression of water in the sucrose solution.