Question

O proba de NaCl contine,ca impuritati 10^-2 moli de clorura unui metal monovalent,MeCl.Stiind ca proba are masa de 6,595 g si contine 59,211% clor, determinati:
a.puritatea probei
b.formula chimica a MeCl

Answer 1

m proba=m NaCl + m impuritati
n impuritati=1/100=0,01moli
n=m/M=> m=n*M
M= A Me + 35,5=A + 35,5
m NaCl=x*M NaCl=58,5x
m impuritati=0,01(A+35,5)=0,01A+0,355
6,595g proba...........100%
35,5x+0,355........59,211%
3550x + 35,5=390,5=>3550x=355=>x=0,1
masa de NaCl=58,5*x=58,5*0,1=5,85g
masa impuritati=6,595-5,85=0,745g
p=mpur*100/mimpur=5,85*100/6,595=88,70%
M impuritati=m/n=0,745/0,01=74,5
A + 35,5=74,5 => A=39 => K
formula lui MeCl: KCl