You are given a gas at 750 torr with a volume of 0.62 L at 25°C. A while later, you measure the temperature and notice a 5°C temperature increase in the system. Assuming there is no pressure change and that this is an ideal gas, what is the new volume?

The initial temperature $T_1 = 25^\circ\text{C}$.

$$T_1 = 25 + 273.15 = 298.15 \, \text{K}$$

The temperature increases by $5^\circ\text{C}$, so the final temperature $T_2 = 25^\circ\text{C} + 5^\circ\text{C} = 30^\circ\text{C}$.

$$T_2 = 30 + 273.15 = 303.15 \, \text{K}$$

Since the pressure is constant, we can use Charles's Law, which states that for a fixed amount of gas at constant pressure, the volume is directly proportional to its absolute temperature:

$$\frac{V_1}{T_1} = \frac{V_2}{T_2}$$

We are given $V_1 = 0.62 \, \text{L}$. We need to find $V_2$.
Rearrange the formula to solve for $V_2$:

$$V_2 = V_1 \times \frac{T_2}{T_1}$$

$$V_2 = 0.62 \, \text{L} \times \frac{303.15 \, \text{K}}{298.15 \, \text{K}}$$ $$V_2 = 0.62 \, \text{L} \times 1.01676$$ $$V_2 \approx 0.63039 \, \text{L}$$

Rounding to two significant figures, the new volume is $0.63 \, \text{L}$.

The final answer is $\boxed{\text{0.63 L}}$.

The given temperature $T = 30^\circ\text{C}$.

$$T = 30 + 273.15 = 303.15 \, \text{K}$$

The given pressure $P = 780 \, \text{torr}$. Convert torr to atmospheres (1 atm = 760 torr):

$$P = 780 \, \text{torr} \times \frac{1 \, \text{atm}}{760 \, \text{torr}} \approx 1.0263 \, \text{atm}$$

The given density $\rho = 3.33 \, \text{g/L}$.
The ideal gas constant $R = 0.08206 \, \text{L} \cdot \text{atm} / (\text{mol} \cdot \text{K})$.

The ideal gas law is $PV = nRT$. We know that the number of moles $n = \frac{m}{M}$, where $m$ is mass and $M$ is molar mass.
Substituting $n$ into the ideal gas law:

$$PV = \frac{m}{M}RT$$

Rearranging to solve for molar mass $M$:

$$M = \frac{mRT}{PV}$$

Since density $\rho = \frac{m}{V}$, we can substitute $\frac{m}{V}$ with $\rho$:

$$M = \frac{\rho RT}{P}$$

$$M = \frac{(3.33 \, \text{g/L})(0.08206 \, \text{L} \cdot \text{atm} / (\text{mol} \cdot \text{K}))(303.15 \, \text{K})}{1.0263 \, \text{atm}}$$ $$M = \frac{82.906}{1.0263} \, \text{g/mol}$$ $$M \approx 80.78 \, \text{g/mol}$$

Rounding to three significant figures, the molar mass is $80.8 \, \text{g/mol}$.

The final answer is $\boxed{\text{80.8 g/mol}}$.