Respuesta :
Answer:
Suppose that you add 29.2 g of an unknown molecular compound to 0.250 kg of benzene, which has a K f of 5.12 oC/m. With the added solute, you find that there is a freezing point depression of 2.78 oC compared to pure benzene. What is the molar mass (in g/mol) of the unknown compound
Explanation:
The mass of nonvolatile solute added is ---- 29.2g
The mass of solvent benzene is ---- 0.250kg = 250g
The Kf value of benzene is ---- 5.12^oC/m.
Depression in the freezing point of the solution is --- 2.78^oC.
What is the molar mass of the unknown solute?
[tex]The depression in freezing point = Kf * molality of the solution\\molality of the solution = \frac{mass of solute}{molar mass of solute}*\frac{1}{mass of solvent in kg}[/tex]
Substitute the given values in this formula to get the molar mass of unknown solvent:
[tex]molality=\frac{29.2g}{M} * \frac{1}{0.250kg} \\depression in freezing point:\\2.78^oC=5.12^oC/m * \frac{29.2g}{M} * \frac{1}{0.250kg} \\\\=>M=5.12^oC/m * \frac{29.2g}{2.78^oC} * \frac{1}{0.250g} \\\\\\=>M=215.1g/mol[/tex]
Hence, the molar mass of unknown solute is --- 215g/mol.
