A gas has density of 1.977 g/dm3 at STP

(a) what is its molecular mass?
(b) what is its density at 77¢XC and 0.95 atm pressure?
(c)If the gas is an oxide and has the formula XO2, what is the atomic mass of X

a.

P*molar mass = density*RT
Substitute and solve for molar mass.

b.
Use the equation in a to solve for density

c.
XO2 = molar mass = ?
2*atomic mass O = 32
Subtract to find atomic mass X.

To answer these questions, we can use the ideal gas law equation:

PV = nRT

Where:
P = Pressure
V = Volume
n = Number of moles
R = Ideal gas constant (0.0821 L·atm/mol·K)
T = Temperature

1) Determining the molecular mass:
To find the molecular mass (molar mass) of the gas, we need to know the number of moles present in a certain volume at STP.

At STP (Standard Temperature and Pressure), the conditions are as follows:
T = 273 K (0°C)
P = 1 atm

Given that the density of the gas at STP is 1.977 g/dm³, we can calculate the molar mass using the following steps:

Step 1: Convert the density to moles per liter (dm³ to L):
1 dm³ = 1 L

Density in g/L = Density in g/dm³

Density in g/L = 1.977 g/dm³ = 1.977 g/L

Step 2: Calculate the number of moles (n) using the ideal gas law equation:

PV = nRT

n = PV / RT

n = (1 atm) x (1 L) / (0.0821 L·atm/mol·K x 273 K)

n = 0.0449 mol

Step 3: Calculate the molar mass (MM) using the formula:

MM = mass / moles

MM = 1.977 g / 0.0449 mol

MM ≈ 44.05 g/mol

Therefore, the molecular mass of the gas is approximately 44.05 g/mol.

2) Determining the density at 77°C and 0.95 atm:
To find the density at a different temperature and pressure, we can apply the ideal gas law equation again.

T = 77°C = 77 + 273 = 350 K
P = 0.95 atm

Step 1: Calculate the number of moles (n) using the ideal gas law equation:

n = PV / RT

n = (0.95 atm) x (1 L) / (0.0821 L·atm/mol·K x 350 K)

n = 0.0402 mol

Step 2: Calculate the density using the formula:

Density = mass / Volume

Mass = n x MM (using the previously calculated molecular mass of 44.05 g/mol)

Mass = 0.0402 mol x 44.05 g/mol

Mass = 1.757 g

Density = mass / Volume

Density = 1.757 g / 1 L

Density = 1.757 g/L

Therefore, the density of the gas at 77°C and 0.95 atm pressure is approximately 1.757 g/L.

3) Determining the atomic mass of X:
The molecular formula of the gas is XO2. From the formula, we can see that the gas consists of one atom of X and two atoms of oxygen (O2).

The molar mass of XO2 is twice the atomic mass of X plus the molar mass of oxygen (O2). From the previously calculated molar mass of the gas (44.05 g/mol), we know that:

MM = 44.05 g/mol

Molar mass of XO2 = 44.05 g/mol

2 x atomic mass of X + molar mass of O2 = 44.05 g/mol

Using the atomic mass of oxygen (O) as 16 g/mol, we can rearrange and solve for the atomic mass of X:

2 x atomic mass of X + 32 g/mol = 44.05 g/mol

2 x atomic mass of X = 44.05 g/mol - 32 g/mol

2 x atomic mass of X = 12.05 g/mol

Atomic mass of X = 12.05 g/mol / 2

Atomic mass of X ≈ 6.025 g/mol

Therefore, the atomic mass of X is approximately 6.025 g/mol.

To answer the given questions, we need to use the Ideal Gas Law equation: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.

(a) To find the molecular mass of the gas, we need to calculate the molar mass. Given the density of the gas at STP, which is 1.977 g/dm³, we can assume that 1 dm³ of the gas would contain a mass of 1.977 grams.

Now we need to convert the mass to moles using the molar mass of the gas. The molar mass of any substance is the mass of one mole of that substance. Let's assume that our gas is represented by the symbol "G" for now.

Using the equation:
moles = mass / molar mass

We substitute the known values:
1.977 g = 1 mole / molar mass

Rearranging the equation:
molar mass = 1 mole / 1.977 g

Thus, the molecular mass of the gas is equal to 1 divided by 1.977, which is approximately 0.506 g/mol.

(b) To find the density of the gas at 77°C and 0.95 atm pressure, we need to apply the Ideal Gas Law equation.

First, we need to convert the given temperature from Celsius to Kelvin:
T(K) = T(°C) + 273.15
T(K) = 77 + 273.15
T(K) = 350.15 K

Now, plug the values into the Ideal Gas Law equation:
PV = nRT

Since we are interested in the density, we can rewrite the equation as follows:
P = (n/V) * RT

We know that the pressure (P) is given as 0.95 atm. The volume (V) is not provided, so we cannot calculate the density directly from the given information. We would need either the volume or the number of moles to proceed.

(c) If the gas is an oxide with the formula XO2, we can determine the atomic mass of "X" by comparing the molar mass calculated in part (a) with the molar mass of XO2. By subtracting the molar mass of O2 (the molecular mass of oxygen, which is 32 g/mol) from the molar mass of XO2, we can find the atomic mass of X.

Molar mass of XO2 = molar mass of X + molar mass of O2

Substituting the values:
0.506 g/mol = (molar mass of X) + 32 g/mol

Rearranging the equation:
molar mass of X = 0.506 g/mol - 32 g/mol

Thus, the atomic mass of X is equal to -31.494 g/mol. Please note that this result is negative, indicating a mistake in the calculations or the given information. Double-checking the calculations is recommended.