Answer :
The weight of the resulting solution is C 94g.
To find the weight of the resulting solution, we need to understand the process of electrolysis. In the electrolysis of AgNO3, Ag is deposited at the cathode. According to Faraday's law, the amount of substance deposited is directly proportional to the quantity of electricity passed through the solution.
First, calculate the moles of Ag deposited using the given information. Since the molar mass of Ag is 108 g/mol and the charge passed is 0.1 F (Faraday), we can use the formula:
\[ \text{moles of Ag} = \frac{\text{charge passed (F)}}{\text{Faraday constant}} = \frac{0.1 \, \text{F}}{1 \, \text{mol/F}} = 0.1 \, \text{mol}\]
Now, since the molar mass of Ag is also given as 108 g/mol, the mass of Ag deposited is:
\[ \text{mass of Ag} = \text{moles of Ag} \times \text{molar mass of Ag} = 0.1 \, \text{mol} \times 108 \, \text{g/mol} = 10.8 \, \text{g}\]
The resulting solution will also contain NO3-, which was originally in the form of AgNO3. Since the charge passed is directly proportional to the moles of Ag deposited, the moles of NO3- are also 0.1 mol.
The molar mass of NO3- is approximately 62 g/mol. Therefore, the mass of NO3- in the solution is:
\[ \text{mass of NO3-} = \text{moles of NO3-} \times \text{molar mass of NO3-} = 0.1 \, \text{mol} \times 62 \, \text{g/mol} = 6.2 \, \text{g}\]
Now, adding the masses of Ag and NO3- together:
\[ \text{Total mass of solution} = \text{mass of Ag} + \text{mass of NO3-} = 10.8 \, \text{g} + 6.2 \, \text{g} = 17 \, \text{g}\]
But this is the mass of the solute (AgNO3) and Ag metal. The question asks for the weight of the resulting solution. Since the weight of water or any solvent is generally much higher compared to the weight of solute, we can assume that the weight of the water hasn't changed significantly. So, the weight of the resulting solution is approximately equal to the weight of the solvent, which is 17 g.
However, it's given that the weight of the resulting solution is 108 g. This means the weight of water (solvent) in the resulting solution is:
\[ \text{Weight of water} = 108 \, \text{g} - 17 \, \text{g} = 91 \, \text{g}\]
But, since the original weight of water was 108 g, and some of it evaporates during electrolysis, the weight of the resulting solution should be less than 108 g. Therefore, the closest option is C. 94g.
Upon electrolyzing a 108g solution of AgNO3 with 0.1 Faraday of electricity, 10.8g of silver is deposited, leaving 97.2g for the final weight of the solution. The options provided do not match this calculated result; hence, the correct answer is 'None of these.'
The question involves the process of electrolyzing a solution of AgNO3. Since the atomic weight of silver is 108, and 1 Faraday (F) of electricity will deposit 108g of silver from AgNO3 solution (1 mole of electrons for each mole of silver), we can determine the final weight of the solution after the electrolysis.
0.1 F of electricity is used, which means that 10.8g (0.1 x 108g) of Ag will be deposited. This mass of silver is removed from the solution. If we had a starting solution weight of 108g, after electrolyzing, we remove 10.8g of Ag from it, leaving us with:
108g (initial weight) - 10.8g (weight of Ag deposited) = 97.2g
However, this result is not among the answer choices provided. As such, it could be a case where none of the answers given (A, B, C, D) directly match the calculated result, indicating a potential error in the question or the answer choices.
