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The density of water at 4°C is $1.0 \times 10^{3} {kgm}^{- 3}$ The volume occupied by one molecule of water is approximately $(N_{A} = 6.0 \times 10^{23})$

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$3.0 \times 10^{- 23} mL$

$6.0 \times 10^{- 23} mL$

$4.0 \times 10^{- 23} mL$

$9.0 \times 10^{- 23} mL$

answer is A.

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Detailed Solution

1g water molecule volume = 1ml

$\frac{1}{18}$ mole water molecule volume = 1ml

1 mol water molecule volume $\begin{aligned} = \frac{1}{\frac{1}{18} \times 6.0 \times 10^{23}} \\ = 3 \times 10^{- 23} ml \end{aligned}$

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