The volume occupied by the molecules contained in 4.5 kg water at STP, if the intermolecular forces vanish away is [2022]

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Q.
The volume occupied by the molecules contained in 4.5 kg water at STP, if the intermolecular forces vanish away is [2022]

1 $5.6 \times 10^{6} m^{3}$

2 $5.6 \times 10^{3} m^{3}$

3 $5.6 \times 10^{- 3} m^{3}$

4 $5.6 m^{3}$

Ans.
(4)

$μ = \frac{Mass of water}{Molecular weight of water}$

$= \frac{4.5 kg}{18 \times 10^{- 3} kg} = 250$

At STP, $T = 273$ $K$ and $P = 10^{5}$ ${N/m}^{2}$

From, $P V = μ R T \Rightarrow V = \frac{μ R T}{P}$

$V = \frac{250 \times 8.3 \times 273}{10^{5}}$

$V = 5.66$ $m^{3}$

Solution

Q.
The volume occupied by the molecules contained in 4.5 kg water at STP, if the intermolecular forces vanish away is [2022]

1 $5.6 \times 10^{6} m^{3}$

2 $5.6 \times 10^{3} m^{3}$

3 $5.6 \times 10^{- 3} m^{3}$

4 $5.6 m^{3}$

Ans.
(4)

$μ = \frac{Mass of water}{Molecular weight of water}$

$= \frac{4.5 kg}{18 \times 10^{- 3} kg} = 250$

At STP, $T = 273$ $K$ and $P = 10^{5}$ ${N/m}^{2}$

From, $P V = μ R T \Rightarrow V = \frac{μ R T}{P}$

$V = \frac{250 \times 8.3 \times 273}{10^{5}}$

$V = 5.66$ $m^{3}$

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Solution

Q. The volume occupied by the molecules contained in 4.5 kg water at STP, if the intermolecular forces vanish away is [2022]

1 5.6×106m3

2 5.6×103m3

3 5.6×10-3m3

4 5.6m3

Ans. (4) μ=Mass of waterMolecular weight of water =4.5 kg18×10-3 kg=250 At STP, T=273 K and P=105 N/m2 From, PV=μRT⇒V=μRTP V=250×8.3×273105 V=5.66 m3

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Q.
The volume occupied by the molecules contained in 4.5 kg water at STP, if the intermolecular forces vanish away is [2022]

1 $5.6 \times 10^{6} m^{3}$

2 $5.6 \times 10^{3} m^{3}$

3 $5.6 \times 10^{- 3} m^{3}$

4 $5.6 m^{3}$

Ans.
(4)

$μ = \frac{Mass of water}{Molecular weight of water}$

$= \frac{4.5 kg}{18 \times 10^{- 3} kg} = 250$

At STP, $T = 273$ $K$ and $P = 10^{5}$ ${N/m}^{2}$

From, $P V = μ R T \Rightarrow V = \frac{μ R T}{P}$

$V = \frac{250 \times 8.3 \times 273}{10^{5}}$

$V = 5.66$ $m^{3}$