Answer to Question #109538 in General Chemistry for Jurni

Question

In the laboratory you are asked to make a 0.150 m chromium(ll) bromide solution using 19.8grams of chromium(ll) bromide. How much water should you add in grams?

Answer 1

n(CrBr2)=m(CrBr2)/Mr(CrBr2) ;

C(CrBr2)=n(CrBr2)/V(H2O)=0.150 M ;

V(H2O)=n(CrBr2)/C(CrBr2)=m(CrBr2)/(C(H2O)*Mr(CrBr2))=19.8/(0.150*212)=0.62264 L=622.64 mL

m(H2O)= ϱ*V(H2O)=1*622.64=622.64 g