3 question was changed to a numbered list, please use \pu for formatting units, see [formatting wiki](https://chemistry.meta.stackexchange.com/a/444) for details
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It is perhaps difficult to comment on the intended semantics of the option C3, but I think that is the most appropriate answer.

AOption 1 and D4 are wrong because of the reasons you cite.

To comment upon Boption 2, we need to know the pH range over which this indicator changes colour. According to Wikipedia, an unequivocal shift from red to yellow occurs from a pH of $3.1$ to a pH of $4.4$. For our case, we would end up with a $\frac{1}{3}\text{M}$$\pu{\frac 13 M}$ solution of HCl$\ce{HCl}$, which, assuming complete dissociation and negligible contribution of $\text{H}^+$ from water, we end up with a pH of around $0.48$ which is well in the no-colour-change region of the pH values.

Regarding Coption 3, the 'more' just refers the water molecules formed as a result of neutralisation reaction, which are additional to the ones already present in either of the mixing solutions. Along with the formation of 'new' NaCl$\ce{NaCl}$ molecules (as in they did not exist bonded to one another before), water molecules are also formed.

$\ce{NaOH + HCl -> NaCl + H2O}$, i.e. table salt and water

It is perhaps difficult to comment on the intended semantics of the option C, but I think that is the most appropriate answer.

A and D are wrong because of the reasons you cite.

To comment upon B, we need to know the pH range over which this indicator changes colour. According to Wikipedia, an unequivocal shift from red to yellow occurs from a pH of $3.1$ to a pH of $4.4$. For our case, we would end up with a $\frac{1}{3}\text{M}$ solution of HCl, which, assuming complete dissociation and negligible contribution of $\text{H}^+$ from water, we end up with a pH of around $0.48$ which is well in the no-colour-change region of the pH values.

Regarding C, the 'more' just refers the water molecules formed as a result of neutralisation reaction, which are additional to the ones already present in either of the mixing solutions. Along with the formation of 'new' NaCl molecules (as in they did not exist bonded to one another before), water molecules are also formed.

$\ce{NaOH + HCl -> NaCl + H2O}$, i.e. table salt and water

It is perhaps difficult to comment on the intended semantics of the option 3, but I think that is the most appropriate answer.

Option 1 and 4 are wrong because of the reasons you cite.

To comment upon option 2, we need to know the pH range over which this indicator changes colour. According to Wikipedia, an unequivocal shift from red to yellow occurs from a pH of $3.1$ to a pH of $4.4$. For our case, we would end up with a $\pu{\frac 13 M}$ solution of $\ce{HCl}$, which, assuming complete dissociation and negligible contribution of $\text{H}^+$ from water, we end up with a pH of around $0.48$ which is well in the no-colour-change region of the pH values.

Regarding option 3, the 'more' just refers the water molecules formed as a result of neutralisation reaction, which are additional to the ones already present in either of the mixing solutions. Along with the formation of 'new' $\ce{NaCl}$ molecules (as in they did not exist bonded to one another before), water molecules are also formed.

$\ce{NaOH + HCl -> NaCl + H2O}$, i.e. table salt and water

2 Showing that water is produced
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It is perhaps difficult to comment on the intended semantics of the option C, but I think that is the most appropriate answer.

A and D are wrong because of the reasons you cite.

To comment upon B, we need to know the pH range over which this indicator changes colour. According to Wikipedia, an unequivocal shift from red to yellow occurs from a pH of $3.1$ to a pH of $4.4$. For our case, we would end up with a $\frac{1}{3}\text{M}$ solution of HCl, which, assuming complete dissociation and negligible contribution of $\text{H}^+$ from water, we end up with a pH of around $0.48$ which is well in the no-colour-change region of the pH values.

Regarding C, the 'more' just refers the water molecules formed as a result of neutralisation reaction, which are additional to the ones already present in either of the mixing solutions. Along with the formation of 'new' NaCl molecules (as in they did not exist bonded to one another before), water molecules are also formed.

$\ce{NaOH + HCl -> NaCl + H2O}$, i.e. table salt and water

It is perhaps difficult to comment on the intended semantics of the option C, but I think that is the most appropriate answer.

A and D are wrong because of the reasons you cite.

To comment upon B, we need to know the pH range over which this indicator changes colour. According to Wikipedia, an unequivocal shift from red to yellow occurs from a pH of $3.1$ to a pH of $4.4$. For our case, we would end up with a $\frac{1}{3}\text{M}$ solution of HCl, which, assuming complete dissociation and negligible contribution of $\text{H}^+$ from water, we end up with a pH of around $0.48$ which is well in the no-colour-change region of the pH values.

Regarding C, the 'more' just refers the water molecules formed as a result of neutralisation reaction, which are additional to the ones already present in either of the mixing solutions. Along with the formation of 'new' NaCl molecules (as in they did not exist bonded to one another before), water molecules are also formed.

It is perhaps difficult to comment on the intended semantics of the option C, but I think that is the most appropriate answer.

A and D are wrong because of the reasons you cite.

To comment upon B, we need to know the pH range over which this indicator changes colour. According to Wikipedia, an unequivocal shift from red to yellow occurs from a pH of $3.1$ to a pH of $4.4$. For our case, we would end up with a $\frac{1}{3}\text{M}$ solution of HCl, which, assuming complete dissociation and negligible contribution of $\text{H}^+$ from water, we end up with a pH of around $0.48$ which is well in the no-colour-change region of the pH values.

Regarding C, the 'more' just refers the water molecules formed as a result of neutralisation reaction, which are additional to the ones already present in either of the mixing solutions. Along with the formation of 'new' NaCl molecules (as in they did not exist bonded to one another before), water molecules are also formed.

$\ce{NaOH + HCl -> NaCl + H2O}$, i.e. table salt and water

1
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It is perhaps difficult to comment on the intended semantics of the option C, but I think that is the most appropriate answer.

A and D are wrong because of the reasons you cite.

To comment upon B, we need to know the pH range over which this indicator changes colour. According to Wikipedia, an unequivocal shift from red to yellow occurs from a pH of $3.1$ to a pH of $4.4$. For our case, we would end up with a $\frac{1}{3}\text{M}$ solution of HCl, which, assuming complete dissociation and negligible contribution of $\text{H}^+$ from water, we end up with a pH of around $0.48$ which is well in the no-colour-change region of the pH values.

Regarding C, the 'more' just refers the water molecules formed as a result of neutralisation reaction, which are additional to the ones already present in either of the mixing solutions. Along with the formation of 'new' NaCl molecules (as in they did not exist bonded to one another before), water molecules are also formed.