Calculating Vapour Pressure for 1L Container with 5mL Water at 20°C

AI Thread Summary
To calculate the vapor pressure of 5 mL of water in a 1 L container at 20°C, first determine the number of moles (n) of water using its mass and molar mass. The mass of 5 mL of water is approximately 4.99 grams, leading to n being 0.277 moles. Using the ideal gas law (PV=nRT), with R as 0.0821 L atm/mol K and the temperature converted to Kelvin (293 K), the equation can be solved. The resulting vapor pressure is approximately 0.0094 atm. This calculation illustrates the relationship between volume, temperature, and vapor pressure in a closed system.
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a 1 L container with 5 mL of water held at 20degcel

I have to figure out the vapour pressure.

I was told to use PV=nRT..but how do I find n?

thanks
 
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First what does n MEAN?
 


To find n (the number of moles), you will need to know the molar mass of water, which is approximately 18 g/mol. Then, you can calculate the number of moles using the following formula:

n = mass of water (in grams) / molar mass of water

In this case, the mass of water is 5 mL, but we need to convert it to grams. The density of water at 20°C is approximately 0.998 g/mL, so 5 mL of water would have a mass of 5 x 0.998 = 4.99 grams.

So, n = 4.99 g / 18 g/mol = 0.277 moles.

Now, you can plug this value of n into the ideal gas law equation, along with the given values for volume (1 L), temperature (20°C = 293 K), and R (the ideal gas constant, which is 0.0821 L atm/mol K):

PV = nRT

(1 L) (x atm) = (0.277 mol) (0.0821 L atm/mol K) (293 K)

x = 0.0094 atm

Therefore, the vapour pressure for a 1 L container with 5 mL of water at 20°C would be approximately 0.0094 atm.
 
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