Theoretical yield is the ideal number, so we'll assume it all reacted without issue.
×
×
×
=56.3gNH↓3
Consider the formation of the three solutionsshown in the table.
1 answer:
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Explanation:
The given reaction is as follows.
Hence, number of moles of NaOH are as follows.
n =
= 0.005 mol
After the addition of 25 ml of base, the pH of a solution is 3.62. Hence, moles of NaOH is 25 ml base are as follows.
n =
= 0.0025 mol
According to ICE table,
Initial: 0.005 mol 0.0025 mol 0 0
Change: -0.0025 mol -0.0025 mol +0.0025 mol
Equibm: 0.0025 mol 0 0.0025 mol
Hence, concentrations of HA and NaA are calculated as follows.
[HA] =
[NaA] =
Now, we will calculate the value as follows.
pH =
=
= 3.42
Thus, we can conclude that of the weak acid is 3.42.
Answer : The molal freezing point depression constant of X is
Explanation : Given,
Mass of urea (solute) = 5.90 g
Mass of X liquid (solvent) = 450.0 g
Molar mass of urea = 60 g/mole
Formula used :
where,
= change in freezing point
= freezing point of solution =
= freezing point of liquid X=
i = Van't Hoff factor = 1 (for non-electrolyte)
= molal freezing point depression constant of X = ?
m = molality
Now put all the given values in this formula, we get
Therefore, the molal freezing point depression constant of X is