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Question: The radius of water molecule having density \( 1gm{{L}^{-1}} \) is: (A) \( 1.925A{}^\circ \) (B...

The radius of water molecule having density 1gmL−11gm{{L}^{-1}} is:
(A) 1.925A∘1.925A{}^\circ
(B) 73.46A∘73.46A{}^\circ
(C) 19.25A∘19.25A{}^\circ
(D) 7.346A∘7.346A{}^\circ

Explanation

Solution

We know that the water is a small bent molecule with the molecular formula H2O.{{H}_{2}}O. It’s found that it’s structure consists of two light hydrogen atoms attached to an oxygen atom. To calculate the number of molecules we hydrogen atoms attached to an oxygen atom. To calculate the number of molecules we 6.02×10236.02\times {{10}^{23}} molecules.

Complete answer:
Water has a dipolar nature, as we know that oxygen has a higher electronegativity than hydrogen, so the side of an oxygen atom has a partial negative charge. A molecule with such a charge difference between water molecules causes water molecules to be attracted to each other and to other polar molecules. This attraction is called hydrogen bonding.
Volume of 1gm1gm water molecules = 1cm3=~1c{{m}^{3}} and Volume of 181 181~ mole of water = 10−6m3=~{{10}^{-6}}{{m}^{3}}
118 \dfrac{1}{18}~ ​ mole of water = 6.023×1023×118=~6.023\times {{10}^{23}}\times \dfrac{1}{18} ​ molecules of water
Volume of 118×6.023×1023 \dfrac{1}{18}\times 6.023\times {{10}^{23~}} molecules   = 10−6m3\,\,=~{{10}^{-6}}{{m}^{3}}
Hence, Volume of one water molecule = 0.2988×10−28m3=~0.2988\times {{10}^{-28}}{{m}^{3}}
Let the radius of water molecule be r=[0.2988×10−28]=[43×π×r3]r=\left[ 0.2988\times {{10}^{-28}} \right]=\left[ \dfrac{4}{3}\times \pi \times {{r}^{3}} \right]
⇒ r=0.1925 nm=1.925A∘\Rightarrow ~r=0.1925~nm=1.925A^\circ
So, the correct answer is option (A).

Additional Information:
By knowing the formula of density and number of molecules, volume can be calculated. From the given density, the relation between mass and volume can be found out. 1 mole of a substance is the molecular mass in grams is equal to the Avogadro’s number. The equation of density is d=mv,d=\dfrac{m}{v}, where d is the density, m is the mass and v is the volume. There could be a confusion between option C and option D because when the power to a number in the denominator will change into opposite when taken to the numerator.

Note:
Remember that we should not get confused in the terms moles and molecules, the number of moles 6.02×10236.02\times {{10}^{23}} molecule of water. To solve any question of finding molecules we need to first find the number of moles present . We must not forget to write the unit after solving any question.