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NEET PHYSICSMedium

The volume occupied by the molecules contained in 4.5 kg water at STP, if the intermolecular forces vanish away is

A

5.6×10−6 m35.6 \times 10^{-6} \text{ m}^3

B

5.6×10−3 m35.6 \times 10^{-3} \text{ m}^3

C

5.6×103 m35.6 \times 10^{3} \text{ m}^3

D

5.6 m35.6 \text{ m}^3

Step-by-Step Solution

Using the ideal gas equation PV=nRTPV = nRT, where n=mass of watermol. wt.=4.5×10318=250 molesn = \frac{\text{mass of water}}{\text{mol. wt.}} = \frac{4.5 \times 10^3}{18} = 250 \text{ moles}. At STP, T=273 KT = 273 \text{ K} and P=105 N/m2P = 10^5 \text{ N/m}^2. Thus, V=nRTP=250×8.3×273105≈5.66 m3V = \frac{nRT}{P} = \frac{250 \times 8.3 \times 273}{10^5} \approx 5.66 \text{ m}^3.

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