Chapter 5, Problem 5.93P

(a)

Interpretation Introduction

Interpretation:

The molar mass of each liquid in all the samples at $70.00°\text{C}$ and $750.0\text{ mL}$ is to be calculated.

Concept introduction:

The ideal gas equation can be expressed as follows:

  $PV=nRT$

Here, $P$ is the pressure, $V$ is the volume, $T$ is the temperature, $n$ is the mole of the gas and $R$ is the gas constant.

The expression to calculate the moles of gas is as follows:

  $\text{Moles of gas}=\frac{\text{Mass of gas}}{\text{Molar mass of gas}}$

(b)

Interpretation Introduction

Interpretation:

From the mass percent of boron in each sample, the molecular formula for each sample is to be determined.

Concept introduction:

The formula to find an amount(mol) is:

  $\text{Amount}\left(\text{mol}\right)=\frac{\text{mass}}{\text{molar mass}}$        (4)

An empirical formula gives the simplest whole number ratio of atoms of each element present in a compound. The molecular formula tells the exact number of atoms of each element present in a compound.

Following are the steps to determine the molecular formula of a compound.

Step 1: Add the molar mass of each element multiplied by its number of atoms present in the empirical formula to obtain the empirical formula mass for the compound.

Step 2: Divide the molar mass of the compound by its empirical formula mass to obtain the whole number. The formula to calculate the whole number multiple is as follows:

  $\text{Whole}-\text{number multiple}=\frac{\text{Molar mass of compound}}{\text{Empirical formula mass}}$        (5)

Step 3: Multiply the whole number with the subscript of each element present in the empirical formula. This gives the molecular formula of the compound.

(c)

Interpretation Introduction

Interpretation:

The molecular formula of sample IV that contains $78.14\text{ \%}$ boron is to be calculated.

Concept introduction:

Effusion is explained as the movement of the gas molecule through a pinhole.

Diffusion can be explained as the mixing of one gas molecule with another gas molecule by random motion.

According to Graham’s law of effusion, the rate of effusion of a gas is inversely proportional to the square root of its molar mass.

The mathematical expression of Graham’s law of effusion is as follows:

  $\frac{\text{Rate of A}}{\text{Rate of B}}=\frac{\sqrt{M_{\text{B}}}}{\sqrt{M_{\text{A}}}}$

Here,

$M_{\text{B}}$ is the molar mass of gas $\text{B}$

$M_{\text{A}}$ is the molar mass of gas $\text{A}$.