Question

The vapor pressure of water is 28.3 mm Hg at 28 °C. What mass of water vapor, in mg, would be present in a vapor volume of 600. mL at 28°C? Selected Answer: B. 16.3 Correct Answer B. 16.3

Answer 1

To find the mass of water vapor in a given vapor volume at a certain temperature, you can use the ideal gas law. Convert the temperature from Celsius to Kelvin, calculate the number of moles using the ideal gas law equation, and then convert moles to grams using the molar mass of water. Finally, convert grams to milligrams. The mass of water vapor in a vapor volume of 600 mL at 28 °C is 5097 mg.

Step-by-step explanation:

To find the mass of water vapor, we can use the ideal gas law equation: PV = nRT, where P represents pressure, V represents volume, n represents the number of moles, R represents the ideal gas constant, and T represents temperature in Kelvin. First, we need to convert the temperature from Celsius to Kelvin by adding 273.15. Then, we can rearrange the equation to solve for n, the number of moles. Once we have the moles, we can use the molar mass of water to convert it to mass in grams. Finally, we can convert grams to milligrams by multiplying by 1000.

Given:

Vapor pressure of water = 28.3 mm Hg
Temperature = 28 °C
Vapor volume = 600 mL

Converting the temperature to Kelvin:

T = 28 °C + 273.15 = 301.15 K

Using the ideal gas law equation to find the number of moles (n):

n = PV / RT

Substituting the values:

n = (28.3 mm Hg * 600 mL) / (62.36 L mm/mol K * 301.15 K)

n = 0.2829 mol

Using the molar mass of water (18.015 g/mol) to find the mass:

mass = n * molar mass

mass = 0.2829 mol * 18.015 g/mol

mass = 5.097 g

Converting grams to milligrams:

mass = 5.097 g * 1000 mg/g

mass = 5097 mg

Therefore, the mass of water vapor in a vapor volume of 600 mL at 28 °C is 5097 mg.

Answer 2

Step-by-step explanation:

The given data is as follows.

P = 28.3 mm =
`(28.3)/(760)`, V = 600 mL =
`600 ml (0.001 L)/(1 ml)` = 0.6 L

R = 0.082 L atm/mol K, T = (28 + 273) K = 301 K

Therefore, according to ideal gas law PV = nRT. Hence, putting the values into this equation calculate the number of moles as follows.

PV = nRT

`(28.3)/(760) * 0.6 L` =
`n * 0.082 L atm/mol K * 301 K`

n =
`9.03 * 10^(-4)` mol

As it is known that number of moles equal to mass divided by molar mass. Hence, mass of water vapor present will be calculated as follows. (molar mass of water is 18 g/mol)

No. of moles =
`\frac{mass}{\text{molar mass}}`

`9.03 * 10^(-4)` mol =
`(mass)/(18 g/mol)`

mass =
`162.5 * 10^(-4)g`

=
`163 * 10^(-4)g` (approx)

Since, 1 g = 1000 mg. Therefore,
`163 * 10^(-4)g` will be equal to
`163 * 10^(-4)g * (10^(3)mg)/(1 g)`

= 16.3 mg

Thus, we can conclude that mass of water vapor present is 16.3 mg.