The density of dichloromethane, an organic solvent, is 1.33 g/mL at 4°C. How many dichloromethane molecules are present in 2.35 mL of dichloromethane at this temperature? (Molar Mass of dichloromethane: 84.93 g/mol)
The correct answer and explanation is :
To determine how many molecules of dichloromethane are present in 2.35 mL of the liquid at 4°C, we need to follow these steps:
Step 1: Convert volume of dichloromethane to mass
Given:
- The density of dichloromethane = 1.33 g/mL
- Volume of dichloromethane = 2.35 mL
We can calculate the mass (m) of dichloromethane using the formula:
$$
\text{Mass} = \text{Density} \times \text{Volume}
$$
Substituting the given values:
$$
\text{Mass} = 1.33 \, \text{g/mL} \times 2.35 \, \text{mL} = 3.1305 \, \text{g}
$$
Step 2: Convert the mass of dichloromethane to moles
Now that we have the mass of dichloromethane, we can convert it to moles using the molar mass. The molar mass of dichloromethane is given as 84.93 g/mol.
To find the number of moles (n), we use the equation:
$$
\text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}}
$$
Substituting the values:
$$
\text{Moles} = \frac{3.1305 \, \text{g}}{84.93 \, \text{g/mol}} = 0.0369 \, \text{mol}
$$
Step 3: Convert moles to molecules
To find the number of molecules, we use Avogadro’s number, which is $6.022 \times 10^{23}$ molecules per mole.
$$
\text{Number of molecules} = \text{Moles} \times \text{Avogadro’s number}
$$
Substituting the values:
$$
\text{Number of molecules} = 0.0369 \, \text{mol} \times 6.022 \times 10^{23} \, \text{molecules/mol}
$$
$$
\text{Number of molecules} = 2.22 \times 10^{22} \, \text{molecules}
$$
Final Answer:
The number of dichloromethane molecules present in 2.35 mL of dichloromethane at 4°C is approximately $2.22 \times 10^{22}$ molecules.
Explanation:
- Density to Mass Conversion: The density tells us how much mass is present in a given volume of the liquid. By multiplying the density (1.33 g/mL) by the volume (2.35 mL), we get the mass of the liquid.
- Mass to Moles Conversion: Using the molar mass of dichloromethane (84.93 g/mol), we can convert the mass into moles. This tells us how many moles of dichloromethane are present in the given volume.
- Moles to Molecules Conversion: Finally, by multiplying the number of moles by Avogadro’s number, we can calculate the number of individual molecules in the sample.
These steps allow us to move from a macroscopic measurement (volume) to a microscopic understanding (molecular count), illustrating the relationship between mass, moles, and molecules.