Question

How many grams of CO(g) are there in 74.5 mL of the gas at 0.933 atm and 30o C?

Answer 1

Answer:

As per the given statement:

$V = 74.5mL$

$P = 0.933 atm$

$T = 30^{\circ} C$

Using ideal gas law equation:

$PV = nRT$

where

P represents the Pressure of a gas

V represents the Volume it occupies

R represents the Universal constant , usually given as:

R = 0.082 atm L/ mol k

First convert Celsius into Kelvin.

T = 273 + 30 = 303 K

Use conversion:

1 L = 1000 mL

V = 74.5 mL = $74.5 \times 10^{-3}$ L

Now substitute the given values we have;

$n = \frac{PV}{RT}$

$n = \frac{0.933 \cdot 74.5 \cdot 10^{-3}}{0.082 \cdot 303}$

Simplify:

n = 0.00279757305

or

n ≈0.0028 moles.

Since carbon monoxide, CO has molar mass of 28.01 g/mol

then;

$m = 0.0028 \times 28.01 = 0.078$ g

Therefore, 0.078 g of CO(g) are there  in 74.5 mL of the gas at 0.933 atm and 30 degree C