A solution (in this experiment solution of NaNO₃) freezes at a lower temperature than does the pure solvent (deionized water). The higher the
solute concentration (sodium nitrate), freezing point depression of the solution will be greater.
Equation describing the change in freezing point:
ΔT = Kf · b · i.
ΔT - temperature change from pure solvent to solution.
Kf - the molal freezing point depression constant.
b - molality (moles of solute per kilogram of solvent).
i - Van’t Hoff Factor.
First measure freezing point of pure solvent (deionized water). Than make solutions of NaNO₃ with different molality and measure separately their freezing points. Use equation to calculate Kf.
Answer:
k = -0.006.
T₀ = 15 °C
Explanation:
Hola.
En este caso, considerando la gráfica mostrada en el archivo adjunto, podemos evidenciar que los datos dados se comportan de manera lineal, por lo que basado en la ecuación, T=k*h+To, podemos calcular la pendiente que basicamente es igual a k, tomando dos puntos en la gráfica:

Además, el valor de la temperatura inicial se puede extraer de la tabla, dado que esta es cuando la altura es 0 m, es decir 15 °C.
¡Saludos!
Answer:
Enthalpy of vaporization = 30.8 kj/mol
Explanation:
Given data:
Mass of benzene = 95.0 g
Heat evolved = 37.5 KJ
Enthalpy of vaporization = ?
Solution:
Molar mass of benzene = 78 g/mol
Number of moles = mass/ molar mass
Number of moles = 95 g/ 78 g/mol
Number of moles = 1.218 mol
Enthalpy of vaporization = 37.5 KJ/1.218 mol
Enthalpy of vaporization = 30.8 kj/mol