Student a performed gravimetric analysis for sulfate in her unknown using the same procedures we did . results of her three trials were 68.6%, 66.2% and 67.1% sulfate. student b analyzed the same unknown his results were 66.7%, 66.6% and 66.5%. the unknown was sodium sulfate. calculate a percent e rror fo r student a and for student b using as the accepted value the theoretical value for sulfate in sodium sulfate based on molar mass . [y ou may use the internet, a textbook, the chemistry c ommunity learning center (cclc) or any other resource for an explanation/description of percent error .] which student , a or b, was more accurate? which student was more precise? explain your answers

The answer is Sodium Sulfate = Na2SO4  

Explanation:

Molar mass of sulfate = 1 (S) + 4 (O) = 1 (32) + 4 (16) = 32 + 64 = 96  

Molar mass of sodium sulfate = 2 (23) + 96 = 46 + 96 = 142  

% of Sulfate = (96/142)*100 = 67.6%  

Percent mistake in Studen A,  

(I) % mistake = (67.6 - 68.6)/67.6 = 1.48  

(ii) % mistake = (67.6 - 66.2)/67.6 = 2.07  

(iii) % mistake = (67.6 - 67.1)/67.6 = 0.74  

For understudy B  

(I) % mistake = (67.6 - 66.7)/67.6 = 1.33  

(ii) % mistake = (67.6 - 66.6)/67.6 = 1.48  

(iii) % mistake = (67.6 - 66.5)/67.6 = 1.63  

Sutdent An is some how exact.  

Understudy B is exact however not precise.

in the given question, the partial pressure is calculated by using the ideal gas equation, that is, pv = nrt

here, p is pressure, v is volume = 25 l, n is number of moles = 0.0104 moles, r is gas constant = -0.0821 l.atm / mol.k

t is temperature = 273 k

p = nrt / v

p = (0.0104 mol × 0.0821 l.atm / mol.k × 273 k) / 25 l

= 0.0093 atm