"resutl" - 1 õppematerjal

Laboratory work no.1 english

m compound ×100 C= m mixture Given mass of the mixture: mmixture=10 g 7,756 g × 100 C NaCl∈ solution= =77,56 10 g 5) Actual percentage of NaCl (given): C NaCl actual=8 0 Calculation of actual mass: 10 g× 8 0 mNaCl actual= =8 g 100 Calculation of the systematic error of the test: ∆=result m−actual m ∆=7,756 g−8 g=−0,244 g resutl m−actual m ∆= × 100 actual m 0,244 g ∆= × 100 =3,1459 7,756 g Conclution Through this experiment I got percentage of NaCl in sand-salt mixture 77,56%. As it was given that percentage of NaCl is 80%, we can consider our result as quite accurate as error was less than 10% (3,14%<10%).