Answer
765.42 torr
Work Step by Step
Let $T_{1} = 155^{o}C + 273 K = 428 K$, $P_{1} = 1200 torr$,
$P_{2} = ?$ $T_{2} = 0^{o}C + 273 = 273K$,
Using Gay Lussac's Law, we have:
$\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}$
$\frac{1200 torr}{428 K} = \frac{P_{2}}{273 K}$
$P_{2} = \frac{273 K}{428 K}$ x 1200 torr
= 765.42 torr
