Question

If the methane contained in 2.50 L of a saturated solution at 25 ∘C was extracted and placed under STP conditions, what volume would it occupy?

Answer 1

116.6 x $10^{ -3$} is the volume of methane contained in 2.50 L of a saturated solution at 25 ∘C that was extracted and placed under STP conditions.

What is an ideal gas equation?

The ideal gas law (PV = nRT) relates to the macroscopic properties of ideal gases. An ideal gas is a gas in which the particles (a) do not attract or repel one another and (b) take up no space (have no volume).

The solubility of methane in water at 25 degrees Celcius is 1.3 x $10^{ -3}$ M. It implies   1.3 x $10^{ -3}$ moles of methane are dissolved in one litre of water.

The number of moles of methane in 4 L of water can be calculated as follows:

Moles of methane = $\frac{1.3 . 10^{ -3} }{1L}$ x$\frac{4L}{1L}$

Moles of methane = 5.2 x $10^{ -3}$

STP refers to standard temperature and pressure. Under STP conditions, the temperature of the substance is  0 degrees celcius and its pressure is 1 atm.

An ideal gas is an imaginary gas comprising of a large number of randomly moving particles and the motion between such articles is considered to be perfectly elastic. The ideal gas equation describes the relationship between pressure, volume, temperature and number of moles of a gas.

The expression for ideal gas equation is as follows:

PV=nRT, where n is the moles and R is the gas constant. Then divide the given mass by the number of moles to get molar mass.

Given data:

P= 1 atm

V= ?

R= $0.082057338 \;L \;atm \;K^{-1}mol^{-1}$

T=273K

n=?

Putting value in the given equation:

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

5.2 x $10^{ -3} = \frac{1 \;atm\; X \;V}{0.082057338 \;L \;atm \;K^{-1}mol^{-1} X 273}$

V= 116.6 x $10^{ -3}$

Hence, 116.6 x $10^{ -3$} is the volume of methane contained in 2.50 L of a saturated solution at 25 ∘C was extracted and placed under STP conditions.

Learn more about the ideal gas here:

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