To determine the molecular weight of the liquid from the given data, you can use the Ideal Gas Law, which is represented by the equation:
\[ PV = nRT \]
where \( P \) is the pressure, \( V \) is the volume, \( n \) is the number of moles of gas, \( R \) is the ideal gas constant, and \( T \) is the temperature in Kelvin.
Given:
- The mass \( m \) of the liquid is 2.75 g.
- The temperature \( T \) is 150 °C, which needs to be converted to Kelvin:
\[ T(K) = T(°C) + 273.15 = 150 + 273.15 = 423.15 K \]
- The volume \( V \) is 255 mL, which needs to be converted to liters since the ideal gas constant \( R \) is usually given in units of liters. So:
\[ V = 255 \text{ mL} \times \frac{1 \text{ L}}{1000 \text{ mL}} = 0.255 \text{ L} \]
- The pressure \( P \) is 0.975 atm.
We need to find the number of moles \( n \) and the molecular weight \( M \). The number of moles \( n \) is related to the mass \( m \) and the molecular weight \( M \) by the equation:
\[ n = \frac{m}{M} \]
We can rearrange the Ideal Gas Law to solve for \( n \):
\[ n = \frac{PV}{RT} \]
Now, plug in the values:
\[ n = \frac{0.975 \text{ atm} \times 0.255 \text{ L}}{(0.0821 \text{ L atm/mol K}) \times 423.15 \text{ K}} \]
Now let's calculate it step by step:
\[ n = \frac{0.975 \times 0.255}{0.0821 \times 423.15} \]
\[ n = \frac{0.248625}{34.745515} \]
\[ n = 0.007158376 \text{ moles} \]
Now that we have the number of moles (\( n \)), we can solve for the molecular weight (\( M \)) using the equation \( n = \frac{m}{M} \), or rearranged to \( M = \frac{m}{n} \):
\[ M = \frac{m}{n} = \frac{2.75 \text{ g}}{0.007158376 \text{ moles}} \]
Now calculate the molecular weight:
\[ M = \frac{2.75}{0.007158376} \]
\[ M = 384.14 \text{ g/mol} \]
So the molecular weight of the liquid is approximately 384.14 g/mol.
