Question

40 UNKNOWN NUMBER Data Trial 2 Trial 1 ২৪.6209 30.744g 30.330j 30.398 2&.256g Mass of the crucible and cover &&.4369 Mass of crucible, cover and sample 7989 Mass of crucible, cover and salt after first heating Mass of crucible, cover and salt after second heating &.99 Mass of crucible, cover and salt after third heating Calculations CaSOu CesOy Chemical formula of anhydrous salt (given) 2.1149 1. 6989 9-689 Mass of hydrate (sample) Mass of anhydrous salt 04169 Mass of water liberated 19-68 90.4401 Percentage H20 in the hydrate go -32% 29-56% Percentage anhydrous salt in the hydrate 0.584 mal J. 14 mol Moles of anhydrous salt in 100 g of hydrate 109 mo Moles of water in 100 g of hydrate Moles of water per mole of anhydrous salt 2 males Average Moles of water per mole of anhydrous salt Casoyn 2H20 Formula of the Hydrate

Answer 1

1. Clearly the result is showing that there is a moles of water per more of casoy - These in no significant weight difference between the first and second heating reading, which confirms that almost all water has been dehydrated in the first heating. The experimental data is showing that 1.95 mole of watch is determined So, it indicates that either theis was a little essor in weighing or more intense heating is required, heating ca504 2H20 - 0. Casou 2. + 2H2O 3. If a hydrate is heated won not strongly heated to remove all watch, the percentage of water will be less as all the watch content is not removed and the weight of compound after heating will show greates • The number of entor mole of water per mote of anhydrous salt will be lower because the weight of water is determined by the mass difference between the sample before and ofter heating, So, if all the water is not removed the sample mars after heating will be greates and the mass of watch will be lower, so, mole of water will also be lower.