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First of all, it is impossible to have $1L$ of liquid water in vapor form in a $1L$ container. The gas molecules would be as close to each other as they were in the liquid form.

However, looking at your last paragraph, it can be inferred what you are actually asking for. I'll consider a very large box instead of a $1L$ box to answer this question.

What does a weighing scale measure?

A weighing scale measures the force applied on the weighing scale's platform by the test object.

Measuring the weight of liquid water

Consider a hypothetical situation where the water exists only in liquid form. There is no vapour pressure. Assume that the box is heavy enough that effects due to atmospheric pressure can be ignored.

The force which the box applies is given by:

$$F = m_{box}g + m_{water}g$$

The scale will read $(m_{box} + m_{water})g$.

Measuring the weight of gaseous water

The gas inside the container applies a pressure on the sides of the container.

Pressure on the top of the container = $P_{top}$

Pressure on the bottom of the container = $P_{top} + \rho gh$

The pressure on the top will help in reducing the force applied by the box on the measuring scale's platform and the pressure applied at the bottom will help to increase the weight applied by the box.

The horizontal forces applied on the sides of the box by the gas will cancel out neatly to give a net horizontal force of $0N$.

The force applied by the box on the scale is given by:

$$F = m_{box}g + (P_{bottom} - P_{top})A$$

$$F = m_{box}g + (P_{top} + \rho gh - P_{top})A$$

$$F = m_{box}g + \rho ghA$$

$$V = Ah$$

$$F = m_{box}g + (\rho V)g = m_{box}g + m_{gas}g$$

If you have the same number of molecules in the gaseous state as there are in the liquid state, $m_{water}$ obtained in the previous case is equal to the $m_{gas}$ obtained in this case.

The scale will read $(m_{box} + m_{water})g$.

If you strictly meant $1L$ of gas and $1L$ of water, then the water will definitely weigh more as it is denser but this question would be very silly if that was the case.

First of all, it is impossible to have $1L$ of liquid water in vapor form in a $1L$ container. It is difficult for liquid form and the gaseous form to occupy the same volume. The gas molecules would be as close to each other as they were in the liquid form.

However, looking at your last paragraph, it can be inferred what you are actually asking for. I'll consider a very large box instead of a $1L$ box to answer this question.

What does a weighing scale measure?

A weighing scale measures the force applied on the weighing scale's platform by the test object.

Measuring the weight of liquid water

Consider a hypothetical situation where the water exists only in liquid form. There is no vapour pressure. Assume that the box is heavy enough that effects due to atmospheric pressure can be ignored.

The force which the box applies is given by:

$$F = m_{box}g + m_{water}g$$

The scale will read $(m_{box} + m_{water})g$.

Measuring the weight of gaseous water

The gas inside the container applies a pressure on the sides of the container.

Pressure on the top of the container = $P_{top}$

Pressure on the bottom of the container = $P_{top} + \rho gh$

The pressure on the top will help in reducing the force applied by the box on the measuring scale's platform and the pressure applied at the bottom will help to increase the weight applied by the box.

The horizontal forces applied on the sides of the box by the gas will cancel out neatly to give a net horizontal force of $0N$.

The force applied by the box on the scale is given by:

$$F = m_{box}g + (P_{bottom} - P_{top})A$$

$$F = m_{box}g + (P_{top} + \rho gh - P_{top})A$$

$$F = m_{box}g + \rho ghA$$

$$V = Ah$$

$$F = m_{box}g + (\rho V)g = m_{box}g + m_{gas}g$$

If you have the same number of molecules in the gaseous state as there are in the liquid state, $m_{water}$ obtained in the previous case is equal to the $m_{gas}$ obtained in this case.

The scale will read $(m_{box} + m_{water})g$.

If you strictly meant $1L$ of gas and $1L$ of water, then the water will definitely weigh more as it is denser but this question would be very silly if that was the case.

First of all, it is impossible to have $1L$ of liquid water in vapor form in a $1L$ container. The gas molecules would be as close to each other as they were in the liquid form.

However, looking at your last paragraph, it can be inferred what you are actually asking for. I'll consider a very large box instead of a $1L$ box to answer this question.

What does a weighing scale measure?

A weighing scale measures the force applied on the weighing scale's platform by the test object.

Measuring the weight of liquid water

Consider a hypothetical situation where the water exists only in liquid form. There is no vapour pressure.

The force which the box applies is given by:

$$F = m_{box}g + m_{water}g$$

The scale will read $(m_{box} + m_{water})g$.

Measuring the weight of gaseous water

The gas inside the container applies a pressure on the sides of the container.

Pressure on the top of the container = $P_{top}$

Pressure on the bottom of the container = $P_{top} + \rho gh$

The pressure on the top will help in reducing the force applied by the box on the measuring scale's platform and the pressure applied at the bottom will help to increase the weight applied by the box.

The horizontal forces applied on the sides of the box by the gas will cancel out neatly to give a net horizontal force of $0N$.

The force applied by the box on the scale is given by:

$$F = m_{box}g + (P_{bottom} - P_{top})A$$

$$F = m_{box}g + (P_{top} + \rho gh - P_{top})A$$

$$F = m_{box}g + \rho ghA$$

$$V = Ah$$

$$F = m_{box}g + (\rho V)g = m_{box}g + m_{gas}g$$

If you have the same number of molecules in the gaseous state as there are in the liquid state, $m_{water}$ obtained in the previous case is equal to the $m_{gas}$ obtained in this case.

The scale will read $(m_{box} + m_{water})g$.

If you strictly meant $1L$ of gas and $1L$ of water, then the water will definitely weigh more as it is denser but this question would be very silly if that was the case.

First of all, it is impossible to have $1L$ of liquid water in vapor form in a $1L$ container. The gas molecules would be as close to each other as they were in the liquid form.

However, looking at your last paragraph, it can be inferred what you are actually asking for. I'll consider a very large box instead of a $1L$ box to answer this question.

What does a weighing scale measure?

A weighing scale measures the force applied on the weighing scale's platform by the test object.

Measuring the weight of liquid water

Consider a hypothetical situation where the water exists only in liquid form. There is no vapour pressure. Assume that the box is heavy enough that effects due to atmospheric pressure can be ignored.

The force which the box applies is given by:

$$F = m_{box}g + m_{water}g$$

The scale will read $(m_{box} + m_{water})g$.

Measuring the weight of gaseous water

The gas inside the container applies a pressure on the sides of the container.

Pressure on the top of the container = $P_{top}$

Pressure on the bottom of the container = $P_{top} + \rho gh$

The pressure on the top will help in reducing the force applied by the box on the measuring scale's platform and the pressure applied at the bottom will help to increase the weight applied by the box.

The horizontal forces applied on the sides of the box by the gas will cancel out neatly to give a net horizontal force of $0N$.

The force applied by the box on the scale is given by:

$$F = m_{box}g + (P_{bottom} - P_{top})A$$

$$F = m_{box}g + (P_{top} + \rho gh - P_{top})A$$

$$F = m_{box}g + \rho ghA$$

$$V = Ah$$

$$F = m_{box}g + (\rho V)g = m_{box}g + m_{gas}g$$

If you have the same number of molecules in the gaseous state as there are in the liquid state, $m_{water}$ obtained in the previous case is equal to the $m_{gas}$ obtained in this case.

The scale will read $(m_{box} + m_{water})g$.

If you strictly meant $1L$ of gas and $1L$ of water, then the water will definitely weigh more as it is denser but this question would be very silly if that was the case.

First of all, it is impossible to have $1L$ of liquid water in vapor form in a $1L$ container. The gas molecules would be as close to each other as they were in the liquid form.

However, looking at your last paragraph, it can be inferred what you are actually asking for. I'll consider a very large box instead of a $1L$ box to answer this question.

What does a weighing scale measure?

A weighing scale measures the force applied on the weighing scale's platform by the test object.

Measuring the weight of liquid water

Consider a hypothetical situation where the water exists only in liquid form. There is no vapour pressure.

The force which the box applies is given by:

$$F = m_{box}g + m_{water}g$$

The scale will read $(m_{box} + m_{water})g$.

Measuring the weight of gaseous water

The gas inside the container applies a pressure on the sides of the container.

Pressure on the top of the container = $P_{top}$

Pressure on the bottom of the container = $P_{top} + \rho gh$

The pressure on the top will help in reducing the force applied by the box on the measuring scale's platform and the pressure applied at the bottom will help to increase the weight applied by the box.

The horizontal forces applied on the sides of the box by the gas will cancel out neatly to give a net horizontal force of $0N$.

The force applied by the box on the scale is given by:

$$F = m_{box}g + (P_{bottom} - P_{top})A$$

$$F = m_{box}g + (P_{top} + \rho gh - P_{top})A$$

$$F = m_{box}g + \rho ghA$$

$$V = Ah$$

$$F = m_{box}g + (\rho V)g = m_{box}g + m_{gas}g$$

If you have the same number of molecules in the gaseous state as there are in the liquid state, $m_{water}$ obtained in the previous case is equal to the $m_{gas}$ obtained in this case.

The scale will read $(m_{box} + m_{water})g$.

If you strictly meant $1L$ of gas and $1L$ of water, then the water will definitely have a higher weight as it is denser but this question would be very silly if that was the case.

First of all, it is impossible to have $1L$ of liquid water in vapor form in a $1L$ container. The gas molecules would be as close to each other as they were in the liquid form.

However, looking at your last paragraph, it can be inferred what you are actually asking for. I'll consider a very large box instead of a $1L$ box to answer this question.

What does a weighing scale measure?

A weighing scale measures the force applied on the weighing scale's platform by the test object.

Measuring the weight of liquid water

Consider a hypothetical situation where the water exists only in liquid form. There is no vapour pressure.

The force which the box applies is given by:

$$F = m_{box}g + m_{water}g$$

The scale will read $(m_{box} + m_{water})g$.

Measuring the weight of gaseous water

The gas inside the container applies a pressure on the sides of the container.

Pressure on the top of the container = $P_{top}$

Pressure on the bottom of the container = $P_{top} + \rho gh$

The pressure on the top will help in reducing the force applied by the box on the measuring scale's platform and the pressure applied at the bottom will help to increase the weight applied by the box.

The horizontal forces applied on the sides of the box by the gas will cancel out neatly to give a net horizontal force of $0N$.

The force applied by the box on the scale is given by:

$$F = m_{box}g + (P_{bottom} - P_{top})A$$

$$F = m_{box}g + (P_{top} + \rho gh - P_{top})A$$

$$F = m_{box}g + \rho ghA$$

$$V = Ah$$

$$F = m_{box}g + (\rho V)g = m_{box}g + m_{gas}g$$

If you have the same number of molecules in the gaseous state as there are in the liquid state, $m_{water}$ obtained in the previous case is equal to the $m_{gas}$ obtained in this case.

The scale will read $(m_{box} + m_{water})g$.

If you strictly meant $1L$ of gas and $1L$ of water, then the water will definitely weigh more as it is denser but this question would be very silly if that was the case.