Respuesta :
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.
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:
0.1336g
Explanation:
Let's bring out what we were given;
Mass of argon = ?
Volume = 75ml = 75 * 10[tex]^{-3}[/tex]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
