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.
The answer is a strike-slip. More specifically a right-lateral strike-slip.
Answer:
0.01144L or 1.144x10^-2L
Explanation:
Data obtained from the question include:
V1 (initial volume) = 20.352 mL
P1 (initial pressure) = 680mmHg
P2 (final pressure) = 1210mmHg
V2 (final volume) =.?
Using the Boyle's law equation P1V1 = P2V2, the volume of the container can be obtained as follow:
P1V1 = P2V2
680 x 20.352 = 1210 x V2
Divide both side by 1210
V2 = (680 x 20.352)/1210
V2 = 11.44mL
Now we need to convert 11.44mL to L in order to obtain the desired result. This is illustrated below:
1000mL = 1 L
11.44mL = 11.44/1000 = 0.01144L
Therefore the volume of the container is 0.01144L or 1.144x10^-2L
The final volume V₂=4.962 L
<h3>Further explanation</h3>
Given
T₁=20 + 273 = 293 K
P₁= 1 atm
V₁ = 4 L
T₂=100+273 = 373 K
P₂=780 torr=1,02632 atm
Required
The final volume
Solution
Combined gas law :
P₁V₁/T₁=P₂V₂/T₂
Input the value :
V₂=(P₁V₁T₂)/(P₂T₁)
V₂=(1 x 4 x 373)/(1.02632 x 293)
V₂=4.962 L
