The volume occupied by the molecules contained in $4.5 ext{ kg}$ water at STP, if the molecular forces vanis

Question

The volume occupied by the molecules contained in $4.5 	ext{ kg}$ water at STP, if the molecular forces vanish away, is:

Accepted Answer

When intermolecular forces vanish, water vapour behaves as an ideal gas. We can calculate the volume using the Ideal Gas Law or the molar volume at STP.

1.  **Calculate Moles ($n$):** 
    Molar mass of water ($H_2O$) = $18 	ext{ g/mol}$.
    Given mass = $4.5 	ext{ kg} = 4500 	ext{ g}$.
    $n = \frac{	ext{Mass}}{	ext{Molar Mass}} = \frac{4500}{18} = 250 	ext{ mol}$.

2.  **Calculate Volume ($V$):**
    At STP (Standard Temperature and Pressure), one mole of an ideal gas occupies $22.4 	ext{ Litres}$ (standard approximation for NEET calculations; older definition).
    $V = n 	imes 22.4 	ext{ L/mol}$
    $V = 250 	imes 22.4 = 5600 	ext{ Litres}$.

3.  **Convert to SI Units ($m^3$):**
    Since $1 	ext{ m}^3 = 1000 	ext{ Litres}$:
    $V = \frac{5600}{1000} 	ext{ m}^3 = 5.6 	ext{ m}^3$.

(Note: Using the modern IUPAC STP molar volume of $22.7 	ext{ L}$ would yield $5.675 	ext{ m}^3$, which is closest to option A).

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