A118.63 g sample of srcl2·6h2o was heated and 34.81 grams of water were recovered from the sample. what percent of the original h2o remains in the sample following this heating?

Percentage mass of water in SrCl₂·6H₂O
= (6 x 18) / (88 + 35.5 x 2 + 6 x 18)
= 40.4%

Mass of water in sample originally₂:
118.63 x 40.4%
= 47.9 grams

Percentage remaining = (47.9 - 34.81) / 47.9 x 100
= 27.3%

Ithink is a because if say increased in electronegativity