Correct Answer: Option B

Explanation

Given: 

mass of water = 18 ; but there are 10 moles of H\(_2\)O = 10 x 18 = 180

mass of Na\(_2\)CO\(_3\).10H\(_2\)O= 286

.: Percentage by mass of water = \(\frac{mass of water}{mass of Na{_2}CO{_3}.10H{_2}O}\) x 100

                                                   = \(\frac{180}{286}\) X 100

                                                   = 62.9%

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