Question

Problema 5.8 va rog!

Answer 1

a)MH3PO4=3AH+AP+4AO=3×1+31+4×16=98g/mol
EgH3PO4=M/3=98/3=32.66g Cn=md/Eg×Vs=5.88/32.66×0.3=5.88/9.798=0.6 Eg/L H3PO4
b) MKMnO4=AK+AMn+4AO=39+55+4×16=158g/mol
EgKMnO4=M/5=158/5=31.6g
Cn=md/Eg×Vs==>Vs=md/Cn/Eg=39.5/1.5/31.6=0.8333 L KMnO4
c) V=1L
Cm=n/Vs ==>n=2 moli H2SO4
md=98×2=196g H2SO4
196g H2SO4................1000mL
T=0.196g/mL.....................1mL
MH2SO4=98g/mol
d) V=1L=1000mL
T=md/V==>md=T×V=0.252g/mL×1000mL=252g
n=md/M=252/63=4 moli HNO3
Cm=n/Vs=4/1=4moli/L
MHNO3=63g/mol