Calculate the amount (in grams) of kcl present in 75.0 ml of 2.10 m kcl
1 answer:
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Explanation:
The given data is as follows.
= 85.0 ml,
= 0.9 M
= 85.0 ml,
= 0.9 M
Hence, number of moles of HCl and KOH will be the same because both the solutions have same volume and molarity.
So, No. of moles = Molarity × Volume
= (as 1 L = 1000 ml so, 85 ml = 0.085 L)
= 0.076 mol
As 1 mole gives 56.2 kJ/mol of heat of neutralization. Hence, calculate the heat of neutralization given by 0.076 moles as follows.
= 4.271 kJ
or, = 4271 J (as 1 kJ = 1000 J)
Therefore, heat released = - heat of gained by calorimeter
Since, it is given that density of the solution is similar to the density of water which is 1 g/ml.
Hence, mass of HCl = density × Volume of HCl
= 1.00 g/ml × 85.0 ml
= 85 g
Similarly, mass of KOH = = density × Volume of HCl
= 1.00 g/ml × 85.0 ml
= 85 g
Hence, total mass of the solution = 85 g + 85 g
= 170 g
Also, q =
4271 J =
0.0773 =
=
Thus, we can conclude that final temperature of the mixed solution is .
Answer:
Both reaction A and reaction B are non spontaneous.
Explanation:
For a spontaneous reaction, change in gibbs free energy () should be negative.
We know, , where T is temperature in Kelvin scale.
Reaction A:
As is positive therefore the reaction is non-spontaneous.
If at a temperature T K , the reaction is spontaneous then-
or,
or,
or,
So at a temperature greater than 350 K, the reaction is spontaneous.
Reaction B:
As is positive therefore the reaction is non-spontaneous.
If at a temperature T K , the reaction is spontaneous then-
or,
or,
or,
So at a temperature greater than -16 K, the reaction is spontaneous.