Scelta multipla
A solution containing 20.0 g of an unknown liquid and 110.0 g of water has a freezing point of -1.32 °C. Given Kf = 1.86 °C/m for water, the molar mass of the unknown liquid is ________ g/mol.
523
views
14. Solutions
Freezing Point Depression
Scelta multipla
Calculate the freezing point of a solution containing 144 g of naphthalene (C10H8) dissolved in 256 g of benzene (C6H6). The freezing point depression constant (Kf) for benzene is 5.12 °C/m. Assume the molar mass of naphthalene is 128.17 g/mol.
A
-1.15 °C
B
-3.45 °C
C
-4.60 °C
D
-2.30 °C
Guida verificata passo dopo passo1
Calculate the number of moles of naphthalene (C10H8) using its mass and molar mass. Use the formula: \( \text{moles of naphthalene} = \frac{\text{mass of naphthalene}}{\text{molar mass of naphthalene}} \).
Determine the molality of the solution. Molality (m) is defined as the number of moles of solute per kilogram of solvent. Use the formula: \( m = \frac{\text{moles of naphthalene}}{\text{mass of benzene in kg}} \).
Apply the freezing point depression formula: \( \Delta T_f = K_f \times m \), where \( \Delta T_f \) is the change in freezing point, \( K_f \) is the freezing point depression constant, and \( m \) is the molality.
Calculate the new freezing point of the solution by subtracting the change in freezing point (\( \Delta T_f \)) from the pure solvent's freezing point. For benzene, the normal freezing point is 5.5 °C.
Compare the calculated freezing point with the given options to determine the correct answer.
Video correlati
Pratica correlata
