$(a)\;4645mL\qquad(b)\;464.5mL\qquad(c)\;4.645mL\qquad(d)\;0.4645mL$

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Given

Initial conditions:

$P_1 = 74 cm$

$T_1 = 20^{\large\circ}C = (20 + 273) = 293 K$

$V_1 = 512 mL$

Final conditions (STP):

$P_2 = 76 cm$

$T_2 = 0^{\large\circ}C = 273 K$

$V_2 = ?$

We know that

$\large\frac{P_1V_1}{T_1} = \large\frac{P_2V_2}{T_2}$

$\large\frac{74\times512}{293} = \large\frac{76\times V_2}{273}$

$V_2 = \large\frac{74\times512\times273}{293\times76}$

= 464.5 mL

Hence answer is (B)

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