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At standard temperature and pressure, a given sample of water vapor occupies a volume of 2.80 L. What is the weight of the water?

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Step-by-step explanation:

The weight of water vapor can be determined by using the ideal gas law, which relates the pressure, volume, and temperature of a gas to the number of moles of gas present. We can then use the molar mass of water to calculate the weight of the water vapor.

At standard temperature and pressure (STP), the pressure is 1 atm and the temperature is 273.15 K. The ideal gas law is expressed as:

PV = nRT

where P is the pressure in atmospheres, V is the volume in liters, n is the number of moles, R is the gas constant (0.08206 L·atm/mol·K), and T is the temperature in Kelvin.

Rearranging this equation to solve for n, we have:

n = PV/RT

Substituting the values we know, we get:

n = (1 atm)(2.80 L) / (0.08206 L·atm/mol·K)(273.15 K)

n = 0.1082 mol

The molar mass of water is 18.01528 g/mol, so the weight of the water vapor is:

weight = n x molar mass

weight = 0.1082 mol x 18.01528 g/mol

weight = 1.952 g

Therefore, the weight of the water vapor is 1.952 g.

User Grzegorz Grabek
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