The answer to this question is theory
The given question is incomplete. The complete question is:
When 136 g of glycine are dissolved in 950 g of a certain mystery liquid X, the freezing point of the solution is 8.2C lower than the freezing point of pure X. On the other hand, when 136 g of sodium chloride are dissolved in the same mass of X, the freezing point of the solution is 20.0C lower than the freezing point of pure X. Calculate the van't Hoff factor for sodium chloride in X.
Answer: The vant hoff factor for sodium chloride in X is 1.9
Explanation:
Depression in freezing point is given by:
= Depression in freezing point
= freezing point constant
i = vant hoff factor = 1 ( for non electrolyte)
m= molality =

Now Depression in freezing point for sodium chloride is given by:
= Depression in freezing point
= freezing point constant
m= molality =


Thus vant hoff factor for sodium chloride in X is 1.9
Answer:
0.74M
Explanation:
Step 1 :
Data obtained from the question.
Initial concentration (C1) = 3M
Initial volume (V2) = 185mL
Final volume (V2) = 750mL
Final concentration (C2) =..?
Step 2:
Determination of the new concentration of the solution.
The new concentration of the solution can be obtained by using the dilution formula as shown below:
C1V1 = C2V2
3 x 185 = C2 x 750
Divide both side by 750
C2 = 3 x 185 / 750
C2 = 0.74M
Therefore, the new concentration of the solution is 0.74M
Answer:
1-C
2-A
3-B
Explanation:
K=[N204]/[NO2]
In the first question the reaction is at equilibrium. When the temperature is decreased, more N2O4 if formed and K increases. When the temperature is increased, more NO2 is formed and K decreases.