Question

A gaseous fl uorinated methane compound has a density of 8.0 g·L1 at 2.81 atm and 300. K. (a) What is the molar mass of the compound? (b) What is the formula of the compound if it is composed solely of C, H, and F? (c) What is the density of the gas at 1.00 atm and 298 K?

Answer 1

Answer:

(a) M = 70.12 g/mol

(b) $CHF_3$

(c) Density = 2.87 g/L

Explanation:

(a)

Using ideal gas equation as:

$PV=nRT$

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 0.0821 L.atm/K.mol

Also,  

Moles = mass (m) / Molar mass (M)

Density (d)  = Mass (m) / Volume (V)

So, the ideal gas equation can be written as:

$PM=dRT$

At STP,  

Pressure = 2.81 atm  

Temperature = 300 K

Density = 8.0 g/L

Applying the equation as:

2.81 atm × M = 8.0 × 0.0821 L.atm/K.mol × 300 K

⇒M = 70.12 g/mol

Molar mass of the compound = 70.12 g/mol

(b)

The gas which corresponds to this gas which contains C, H and F is $CHF_3$

As, 12 + 1 + 3 × 19 = 70 g/mol which corresponds to the above mass

(c)

Using

$PM=dRT$

P = 1.00 atm

R = 0.0821 L.atm/K.mol

M = 70.12 g/mol

T = 298 K

So,

$1\times 70.12=d\times 0.0821\times 298$

Thus, density = 2.87 g/L