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The radius of water molecule having density 1g mL^-1 is

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19.25 $\overset{o}{A}$

1.925 $\overset{o}{A}$

73.46 $\overset{o}{A}$

7.346 $\overset{o}{A}$

answer is A.

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Detailed Solution

.Volume of 1gm water molecules= 1cm³

Volume of 181 mole of water= 10⁻⁶m³

181 mole of water= 6.023×1023×181 molecules of water

Volume of 181×6.023×1023 molecules= 10⁻⁶m³

Hence, Volume of one water molecule= 0.2988×10⁻²⁸m³

Let. the radius of water molecule be r

0.2988×10⁻²⁸=34×π×r³

⇒ r=0.1925 nm=1.925A˚

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