1) a 1,084g sample of hydrate is heated, leaving 0,762g ofanhydrous residue. a)

Question

1) a 1,084g sample of hydrate is heated, leaving 0,762g ofanhydrous residue. a) how many grams of water were driven off?b)how many moles of water were driven off?c) what is the percent massof water in the hydrate? 2) in the first trial of part A of the procedure, the ratio ofmoles H20/moles CaSO4 is found to be 1,93. in the second trial theratio was found to be 1,98. calculate the relative averagedeviation of the two trials. should the student run anothertrial? 1) a 1,084g sample of hydrate is heated, leaving 0,762g ofanhydrous residue. a) how many grams of water were driven off?b)how many moles of water were driven off?c) what is the percent massof water in the hydrate? 2) in the first trial of part A of the procedure, the ratio ofmoles H20/moles CaSO4 is found to be 1,93. in the second trial theratio was found to be 1,98. calculate the relative averagedeviation of the two trials. should the student run anothertrial?

Explanation / Answer

(1)mass of hydrated sample , m = 1.084 g mass of the anhydrous residue , m' = 0.762 g So, mass of water which is lost during heating is , m'' = m-m'                                                                                 =0.322g                    -----(a) Molar mass of H2O is = 2 * 1 + 16 = 18 g No. of moles , n = mass / Molar mass                          = 0.322 / 18                          = 0.01788mol                                                    -----(b) % mass of water = ( mass of water / mass of sample) * 100                           = ( 0.322 / 1.084 ) * 100                            =29.7

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