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.
I am pretty sure it is 4.002602
Answer:
The empirical formula is N2 CH4 O
Explanation:
The equation is
W = C/F
W= 3.00 x 10^8 m/sec
——————————
6.165 x 10^14 Hz
W= 4.87 x 10^-7 m
Energy is
E=hF
E= (6.626 x 10^-34 Jxsec )(6.165 x 10^14 Hz)
E= 4.085 x 10^-19 J
