Question

The molecular weight of a gas which diffuses through a porous plug of \(1 / 6\) th of the speed of hydrogen under identical conditions is (1) 27 (2) 72 (3) 36 (4) 48

Short answer

The molecular weight of the gas is 72 g/mol.

Step-by-step solution

Recall Graham's Law

Graham's law states that the rate of diffusion of a gas is inversely proportional to the square root of its molar mass: \ \[ \text{Rate of diffusion} \propto \frac{1}{\sqrt{M}} \ \] Where \ M \ is the molar mass.

Set Up the Ratio

Given that the gas diffuses at \ \frac{1}{6}\ \ the speed of hydrogen, we can write: \ \[ \frac{r_{\text{gas}}}{r_{\text{H}_2}} = \frac{1}{6} \ \] where \ r_{\text{gas}} \ and \ r_{\text{H}_2} \ are the rates of diffusion of the gas and hydrogen respectively.

Substitute into Graham's Law

Using Graham's law, we know \ \[ \frac{r_{\text{gas}}}{r_{\text{H}_2}} = \frac{ \sqrt{M_{\text{H}_2}}} { \sqrt{M_{\text{gas}}}} \ \] \ \ Rearranging the equation, we have: \ \[ \frac{1}{6} = \frac{\sqrt{M_{\text{H}_2}}}{ \sqrt{M_{\text{gas}}}} \ \]

Solve for the Molar Mass

Square both sides of the equation to get rid of the square roots: \ \[ \left( \frac{1}{6} \right)^2 = \ \frac{M_{\text{H}_2}}{M_{\text{gas}}} \ \] \ \[ \frac{1}{36} = \ \frac{M_{\text{H}_2}}{M_{\text{gas}}} \ \] \ \ Hydrogen has a molar mass \ (M_{\text{H}_2}) = 2 \text{ g/mol}, so we can substitute this into the equation: \ \[ \frac{1}{36} = \ \frac{2}{M_{\text{gas}}} \ \]

Isolate the Gas Molar Mass

Rearrange to isolate \ M_{\text{gas}}: \ \[ M_{\text{gas}} = 36 \times 2 = 72 \ \text{ g/mol} \ \]

Key concepts

molecular weight

Molecular weight, also called molecular mass, is a fundamental concept in chemistry. It's the weight of a single molecule of a substance and is usually measured in atomic mass units (amu). To find the molecular weight, you must add up the atomic weights of all the atoms in the molecule. For example, water (H2O) consists of two hydrogen atoms (each about 1 amu) and one oxygen atom (about 16 amu). So, the molecular weight of water is approximately 18 amu. Molecular weight is crucial in understanding many chemical behaviors, including how gases diffuse. Knowing the molecular weight allows you to apply Graham's Law effectively and predict how fast different gases will spread out under similar conditions.

rate of diffusion

The rate of diffusion measures how quickly molecules spread from an area of high concentration to an area of low concentration. Graham's Law specifically deals with the diffusion rates of gases. According to this law, the rate of diffusion of a gas is inversely proportional to the square root of its molar mass. In simpler terms, lighter gases (with lower molar masses) diffuse faster than heavier ones. You often compare the rates of two gases by setting up a ratio. For instance, if Gas A diffuses faster than Gas B, then \(\frac{r_A}{r_B}\) would be greater than 1. Using this understanding and Graham's Law, you can predict how quickly one gas will diffuse compared to another.

molar mass calculation

Calculating molar mass is easy if you know the molecular formula of a substance. The molar mass is the mass of one mole of a substance measured in grams per mole (g/mol). Let's take the example of Carbon Dioxide (CO2). Carbon has an atomic mass of about 12 g/mol and Oxygen has an atomic mass of about 16 g/mol. Since CO2 has one carbon atom and two oxygen atoms, its molar mass will be \(12 + 2 \times 16 = 44\) g/mol. When applying this to Graham's Law, molar mass helps you determine the rate at which gases will diffuse. The steps you follow usually involve finding the ratio of diffusion rates and then solving for the unknown molar mass by rearranging the formula so that you can isolate it.