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ilu molowy jest kwas siarkowy bedacy o stezeniu 80 % i gestosci 1,75g/cm3
mH2SO4 = 98g/mol
100g - 80g
<ro>= m/V
V= m/<ro>
V= 100g/1,75g/cm3 = 57 cm3
57cm3 - 80g
1000 cm3 - x
x = 1403g
98g - 1 mol
1403g - x
x = 14 mol
Odp. Stężenie molowe wynosi 14 mol/dm3