State and explain Graham's law of diffusion. 0.23gm of a volatile liquid occupies 126.4 cc at 27°C and 760 mm pressure calculate the molecular weight of the liquid. How many molecules of the liquid…

Graham's Law of Diffusion

Statement and Explanation

Graham's Law of Diffusion (or Effusion) states that the rate of diffusion or effusion of a gas is inversely proportional to the square root of its molecular weight (or density). Mathematically, it is expressed as:

carbon dioxidecarbon dioxideethaneethane

$$\frac{r_1}{r_2}=\sqrt{\frac{M_2}{M_1}}\text{or}\frac{r_1}{r_2}=\sqrt{\frac{d_2}{d_1}}$$

where:

• $r_1$ and $r_2$ are the rates of diffusion/effusion of two gases,
• $M_1$ and $M_2$ are their molecular weights,
• $d_1$ and $d_2$ are their densities.

Key Points:

1. Diffusion refers to the spontaneous mixing of gas molecules due to random motion.
2. Effusion is the escape of gas molecules through a tiny hole (e.g., a pinhole).
3. Inverse proportionality: Lighter gases diffuse/effuse faster than heavier ones.
4. Applications:
   • Separation of gases (e.g., uranium enrichment via effusion).
   • Estimating molecular weights of unknown gases.

---

Calculation of Molecular Weight

Given Data:

• Mass of liquid ($m$) = 0.23 g
• Volume ($V$) = 126.4 cc = 0.1264 L
• Temperature ($T$) = 27°C = 300 K (273 + 27)
• Pressure ($P$) = 760 mm Hg = 1 atm

Step 1: Use the Ideal Gas Law to Find Moles ($n$)

The ideal gas equation is: $$PV=nRT$$ where:

• $R$ = 0.0821 L·atm·K⁻¹·mol⁻¹ (gas constant),
• $n$ = number of moles of gas.

Rearranged to solve for $n$: $$n=\frac{PV}{RT}$$

Substitute the values: $$n=\frac{(1\text{ atm})(0.1264\text{ L})}{(0.0821\text{ L⋅atm⋅K⁻¹⋅mol⁻¹})(300\text{ K})}$$ $$n=\frac{0.1264}{24.63}\approx 0.00513\text{ mol}$$

Step 2: Calculate Molecular Weight ($M$)

Molecular weight is defined as: $$M=\frac{\text{Mass}}{\text{Moles}}=\frac{m}{n}$$

Substitute the values: $$M=\frac{0.23\text{ g}}{0.00513\text{ mol}}\approx 44.83\text{ g/mol}$$

Molecular weight of the liquid ≈ 44.83 g/mol *(Rounded to 45 g/mol for simplicity, likely ethane (C₂H₆) or carbon dioxide (CO₂).)*

CO₂ (Molecular Weight = 44 g/mol)CO₂ (Molecular Weight = 44 g/mol)

---

Number of Molecules in 0.23 g

Step 1: Use Avogadro’s Number

Avogadro’s number ($N_A$) = $6.022\times 10^{23}$ molecules/mol.

Number of molecules ($N$) is given by: $$N=n\times N_A$$

Substitute $n=0.00513$ mol: $$N=(0.00513\text{ mol})\times (6.022\times 10^{23}\text{ molecules/mol})$$ $$N\approx 3.09\times 10^{21}\text{ molecules}$$

Number of molecules in 0.23 g ≈ $3.09\times 10^{21}$ molecules.

---

Verification of Assumptions

1. Volatile Liquid as Gas: The problem states the liquid is volatile, implying it behaves as an ideal gas under given conditions (low pressure, moderate temperature).
2. Ideal Gas Law Applicability: Valid for most gases at STP (Standard Temperature and Pressure). Minor deviations may occur for highly polar or large molecules, but negligible here.
3. Molecular Weight: The calculated value (45 g/mol) matches common gases like CO₂ (44 g/mol) or C₂H₆ (30 g/mol). The slight discrepancy could be due to rounding or experimental error.

---

C₂H₆ (Molecular Weight = 30 g/mol)C₂H₆ (Molecular Weight = 30 g/mol)