A current of 9.65 amp is passed through the aqueous solution NaCl using suitable electrodes for 1000 s. The amount of NaOH formed during electrolysis is

4.0 g

To determine the amount of NaOH formed during electrolysis, we need to calculate the number of moles of electrons that have been transferred in the electrolysis process. From there, we can use the stoichiometry of the reaction to find the amount of NaOH formed.

First, let's calculate the number of moles of electrons transferred:

Given:
Current (I) = 9.65 A
Time (t) = 1000 s

Using the formula:
\[Q = I \cdot t\]

where \(Q\) is the electric charge in Coulombs (C), we can find the value of \(Q\):

\[Q = 9.65 \, \text{A} \cdot 1000 \, \text{s} = 9650 \, \text{C}\]

Since 1 mole of electrons corresponds to 1 Faraday (F) of charge, and 1 Faraday is equal to 96500 C, we can determine the number of moles of electrons transferred:

\[\text{Moles of electrons} = \frac{Q}{96500 \, \text{C/mol}} = \frac{9650 \, \text{C}}{96500 \, \text{C/mol}} = 0.1 \, \text{mol}\]

The balanced chemical equation for the electrolysis of NaCl is:

\[2\text{NaCl(aq)} + 2\text{H}_2\text{O(l)} \rightarrow 2\text{NaOH(aq)} + \text{H}_2(g) + \text{Cl}_2(g)\]

According to the stoichiometry of the reaction, 2 moles of NaOH are formed per mole of electrons transferred. Therefore, the number of moles of NaOH formed is also 0.1 mol.

To convert the moles of NaOH to grams, we need to use the molar mass of NaOH, which is approximately 40 g/mol.

\[\text{Mass of NaOH} = \text{Moles of NaOH} \times \text{Molar mass of NaOH} = 0.1 \, \text{mol} \times 40 \, \text{g/mol} = 4 \, \text{g}\]

Therefore, the amount of NaOH formed during electrolysis is (2) 4.0 g.