## Chemistry: Molecular Approach (4th Edition)

$V_{1}$ = 1.55 mL $T_{1}$ = (95.3+273.15) K = 368.45 K $T_{2}$ = 273.15 K $\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}$ ( Charles' law ) Then, $V_{2}$ = $\frac{V_{1}T_{2}}{T_{1}} = \frac{1.55 mL\times273.15K}{368.45K}$ = 1.15 mL