‘x’ g of molecular oxygen (O2) is mixed with 200 g of neon (Ne). The total pressure of the non-reactive mixture of O2 and Ne in the cylinder is 25 bar. The partial pressure of Ne is 20 bar at the same temperature and volume. The value of ‘x’ is ___.
[Given : Molar mass of O2 = 32 g mol–1 and Molar mass of Ne = 20 g mol–1]
[Given : Molar mass of O2 = 32 g mol–1 and Molar mass of Ne = 20 g mol–1]
Correct Answer: 80
Approach Solution - 1
To solve this problem, we employ the ideal gas law and Dalton's Law of Partial Pressures. Given that the total pressure and partial pressure of neon are provided, we can find the partial pressure of molecular oxygen in the mixture.
- First, use Dalton's Law to find the partial pressure of O2. Dalton's Law states that the total pressure (Ptotal) is the sum of the partial pressures of the gases: Ptotal=P(O2)+P(Ne).
- We have Ptotal=25 bar and P(Ne)=20 bar.
- Thus, P(O2)=Ptotal−P(Ne)=25 bar−20 bar=5 bar.
- Next, we calculate the moles of Ne using the ideal gas law: PV=nRT, where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is temperature.
- We don't need exact values for V, R, and T, since they will cancel out, considering both gases are in the same conditions.
- For Ne: P(Ne)V=n(Ne)RT implies n(Ne)=P(Ne)V/RT.
- Given the weight of Ne is 200 g and its molar mass is 20 g/mol, n(Ne)=200 g / 20 g/mol=10 mol.
- Since we have determined the partial pressure of O2 and know the relationship with moles, we calculate the moles of O2 similarly: n(O2)=P(O2)V/RT.
- Using the ratio of moles and pressures (because V, R, and T cancel out), n(O2)=P(O2)/P(Ne) * n(Ne)=5 bar / 20 bar * 10 mol=2.5 mol.
- Convert moles to grams for oxygen: mass of O2=n(O2) * molar mass of O2=2.5 mol * 32 g/mol=80 g.
The value of ‘x’, which is the mass of O2, is 80 g. It falls within the range 80 to 80, confirming correctness.
Approach Solution -2
\(P_{O_2}=25−20=5\) bar
\(P_{O_2}=X_{O_2}×P_{Total}\)
\(\frac {5}{25}=\frac {n_{O2}}{n_{O_2}+n_{Ne}}\)
\(\frac 15=\frac {\frac {x}{32}}{\frac {x}{32}+\frac {200}{20}}\)
\(⇒\frac {x}{32}+10=\frac {5x}{32}\)
\(⇒\frac x8=10\)
\(⇒x=80\) gm
So, the answer is \(80\).
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Concepts Used:
Solutions
A solution is a homogeneous mixture of two or more components in which the particle size is smaller than 1 nm.
For example, salt and sugar is a good illustration of a solution. A solution can be categorized into several components.
Types of Solutions:
The solutions can be classified into three types:
- Solid Solutions - In these solutions, the solvent is in a Solid-state.
- Liquid Solutions- In these solutions, the solvent is in a Liquid state.
- Gaseous Solutions - In these solutions, the solvent is in a Gaseous state.
On the basis of the amount of solute dissolved in a solvent, solutions are divided into the following types:
- Unsaturated Solution- A solution in which more solute can be dissolved without raising the temperature of the solution is known as an unsaturated solution.
- Saturated Solution- A solution in which no solute can be dissolved after reaching a certain amount of temperature is known as an unsaturated saturated solution.
- Supersaturated Solution- A solution that contains more solute than the maximum amount at a certain temperature is known as a supersaturated solution.