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Lelechka [254]
3 years ago
15

If I add water to 100mL of a 0.15 M NaOH solution until the final volume is 150mL, what will the molarity of the diluted solutio

n be?
Chemistry
1 answer:
asambeis [7]3 years ago
8 0

Answer: The molarity of the diluted solution be 0.10 M

Explanation:

According to the dilution law,

C_1V_1=C_2V_2

where,

C_1 = concentration of concentrated solution = 0.15 M

V_1 = volume of concentrated solution = 100 ml

C_2 = concentration of diluted solution= ?

V_2 = volume of diluted solution= 150 ml

Putting the values we get:

0.15\times 100=M_2\times 150

M_2=0.10M

Thus the molarity of the diluted solution be 0.10 M

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The cell potential of a redox reaction occurring in an electrochemical cell under any set of temperature and concentration condi
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Answer : The actual cell potential of the cell is 0.47 V

Explanation:

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The balanced two-half reactions will be,

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The expression for reaction quotient will be :

Q=\frac{[Zn^{2+}]}{[Ni^{2+}]}

In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted.

Now put all the given values in this expression, we get

Q=\frac{(0.0141)}{(0.00104)}=13.6

The value of the reaction quotient, Q, for the cell is, 13.6

Now we have to calculate the actual cell potential of the cell.

Using Nernst equation :

E_{cell}=E^o_{cell}-\frac{RT}{nF}\ln Q

where,

F = Faraday constant = 96500 C

R = gas constant = 8.314 J/mol.K

T = room temperature = 316 K

n = number of electrons in oxidation-reduction reaction = 2 mole

E^o_{cell} = standard electrode potential of the cell = 0.51 V

E_{cell} = actual cell potential of the cell = ?

Q = reaction quotient = 13.6

Now put all the given values in the above equation, we get:

E_{cell}=0.51-\frac{(8.314)\times (316)}{2\times 96500}\ln (13.6)

E_{cell}=0.47V

Therefore, the actual cell potential of the cell is 0.47 V

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