a 2.00-liter sample of a gas has a mass of 1.80 grams at STP. What is the density , in grams per liter , of this gas at Stp?
0.900
density = mass/volume.
To determine the density of a gas at STP (Standard Temperature and Pressure), we need to use the ideal gas law, which states:
PV = nRT
Where:
P = pressure
V = volume
n = number of moles
R = ideal gas constant
T = temperature
At STP, the conditions are a pressure of 1 atmosphere (atm) and a temperature of 273.15 Kelvin (K).
Given:
Volume (V) = 2.00 liters
Mass (m) = 1.80 grams
First, we need to determine the number of moles (n) of the gas using the equation:
n = m / M
Where:
M = molar mass of the gas
To calculate the molar mass, we need to know the identity of the gas. Let's assume the gas is an ideal gas with a molar mass of X grams/mol.
n = 1.80 g / X g/mol
Now, we can rearrange the ideal gas law formula to solve for density (d):
d = m / V
To express the density in grams per liter (g/L), we need to convert grams into moles first:
d = n * (M / V)
Substituting the values we have into our equation:
d = (1.80 g / X g/mol) * (X g/mol) / 2.00 L
Simplifying the equation:
d = 1.80 / 2.00 g/L
Therefore, the density of the gas at STP is 0.90 grams per liter (g/L).
To calculate the density of the gas at STP, we need to use the formula:
Density = Mass / Volume
Given information:
Mass of the gas = 1.80 grams
Volume of the gas = 2.00 liters
We can substitute these values into the formula:
Density = 1.80 grams / 2.00 liters
Now, we can divide 1.80 grams by 2.00 liters to get the density of the gas:
Density = 0.90 grams per liter
Therefore, the density of the gas at STP is 0.90 grams per liter.