Question

Calculate the radius of a water molecule assuming to be spherical (density $ =1\text{ }g\text{ }m{{L}^{-1}} $ ).

• A. $ 1.925\times {{10}^{-8}}cm $
• B. $ 1.925\times {{10}^{-9}}cm $
• C. $ 1.925\times {{10}^{-10}}cm $
• D. $ 1.925\times {{10}^{-11}}cm $

Answer 1

Answer: (A)

Mass of one molecule $ =\frac{18}{6.02\times {{10}^{23}}}g $ Volume of one molecule $ =\frac{18g}{6.02\times {{10}^{23}}\times 1} $ $ =3.0\times {{10}^{-23}} $ $ \frac{4}{3}\pi {{r}^{3}}=3.0\times {{10}^{-23}} $ $ {{r}^{3}}=\frac{3\times 3.0\times {{10}^{-23}}\times 7}{4\times 22} $ $ r=1.925\times {{10}^{-8}}cm $