Calculate the vapor pressure of pure B using Raoult's Law, given moles of A and B, total pressure, and vapor pressure of A.

Number of moles of A ($n_A$) = $0.650 \text{ mol}$
Number of moles of B ($n_B$) = $0.250 \text{ mol}$

$$n_{total} = n_A + n_B = 0.650 \text{ mol} + 0.250 \text{ mol} = 0.900 \text{ mol}$$

$$X_A = \frac{n_A}{n_{total}} = \frac{0.650 \text{ mol}}{0.900 \text{ mol}} = 0.7222$$ $$X_B = \frac{n_B}{n_{total}} = \frac{0.250 \text{ mol}}{0.900 \text{ mol}} = 0.2778$$

The total vapor pressure of the solution ($P_{total}$) at its normal boiling point is $1 \text{ atm}$.
The vapor pressure of pure A ($P_A^\circ$) is $0.700 \text{ atm}$.
Raoult's Law states:

$$P_{total} = X_A \cdot P_A^\circ + X_B \cdot P_B^\circ$$

Substitute the known values:

$$1.000 \text{ atm} = (0.7222 \times 0.700 \text{ atm}) + (0.2778 \times P_B^\circ)$$

$$1.000 \text{ atm} = 0.50554 \text{ atm} + 0.2778 \cdot P_B^\circ$$

Subtract $0.50554 \text{ atm}$ from both sides:

$$1.000 \text{ atm} - 0.50554 \text{ atm} = 0.2778 \cdot P_B^\circ$$ $$0.49446 \text{ atm} = 0.2778 \cdot P_B^\circ$$

Divide by 0.2778:

$$P_B^\circ = \frac{0.49446 \text{ atm}}{0.2778} = 1.780 \text{ atm}$$

The vapor pressure of pure B at this temperature is $\text{1.780 atm}$.