Question

What is the mass of argon gas in a 75.0 mL volume at STP

Answer 1

Assuming that STP = 0°C, and 1 atm pressure: 
molar volume is 22.414 L = 22 414 cm^3 

75 mL = 1*75/22 414 moles = 0.0033461 moles. 
mole mass = 39.9 g ---> 
0.0033461 moles = 0.0033461*39.9 g = 0.13351 g <--- ans. 

Answer 2

Answer:

0.1336g

Explanation:

Let's bring out what we were given;

Mass of argon = ?

Volume = 75ml = 75 * 10$^{-3}$L = 0.075L

STP (Standard temperature and Pressure)

Standard pressure (P) = 1 atm

Standard Temperature (T) = 273K

Molar mass of Argon = 39.948g/mol

We can use the formula below to calculate the mass of argon;

Mass = No of moles * Molar mass

But we have to find the value no. of moles (n).

From ideal gas equation, we know that PV =nRT

where R = gas constant = 0.0821 L*atm/(mol*K)

n = PV/RT

n = (1 * 75) / (0.0821 * 273)

n = 0.075 / 22.4133

n = 0.003346 moles

Now we can use the molar mass formular to calculate the mass.

Mass = No of moles * Molar mass

Mass = 0.003346 * 39.948 = 0.1336g