We are given the following data:
First, we calculate the number of moles of Q formed:
Moles of Q = \( \frac{\text{Mass of Q}}{\text{Molar mass of Q}} \)
Moles of Q = \( \frac{40 \, \text{g}}{40 \, \text{g/mol}} = 1 \, \text{mol} \)
Since one mole of Q produces one mole of Cl2, we can now calculate the volume of Cl2 using the ideal gas law:
PV = nRT
Substituting the known values:
\( (1 \, \text{atm})(V) = (1 \, \text{mol})(0.082 \, \text{L atm mol}^{-1} \text{K}^{-1})(298 \, \text{K}) \)
V = \( \frac{(1)(0.082)(298)}{1} \)
V = 12.1 L
Thus, the volume of Cl2 formed is 12.1 litres.
If the molar conductivity ($\Lambda_m$) of a 0.050 mol $L^{β1}$ solution of a monobasic weak acid is 90 S $cm^{2} mol^{β1}$, its extent (degree) of dissociation will be:
[Assume: $\Lambda^0$ = 349.6 S $cm^{2} mol^{β1}$ and $\Lambda^0_{\text{acid}}$ = 50.4 S$ cm^{2} mol^{β1}$]