Koliko je vode potrebno dodati u 50g 20% rastvora neke supstance da bi se dobio rastvor masenogg udjela ω=0.04?
mr1=50g w1=0.2 a w2=0.04
ms1=w1*mr1=0.2*50=10 g
ms2=ms1=10g
mr2=ms2/w2=10/0.04=250 g
m(vode)=mr2-mr1=250-50=200 g
powered by Drupal
mr1=50g w1=0.2 a
mr1=50g w1=0.2 a w2=0.04
ms1=w1*mr1=0.2*50=10 g
ms2=ms1=10g
mr2=ms2/w2=10/0.04=250 g
m(vode)=mr2-mr1=250-50=200 g