Koliki je maseni udeo NaCl u uzorku NaCl kada je za titraciju 0.2000g soli potrebno 33.52 cm3 rastvora AgNO3 koncentracije 0.1000 ,ol/dm3 [resenje 97.95%]
n(AgNO3)=33.52cm^3*0.1mol/dm^3=0.003352mol
AgNO3+NaCl-->NaNO3+AgCl
n(NaCl)=n(AgNO3)=0.003352mol
m(NaCl)=n(NaCl)*58.5g/mol=0.195975g
p=0.195975g/2.000g=97.9875%
The goal of life is living in agreement with nature.
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n(AgNO3)=33.52cm^3*0.1mol/dm^
n(AgNO3)=33.52cm^3*0.1mol/dm^3=0.003352mol
AgNO3+NaCl-->NaNO3+AgCl
n(NaCl)=n(AgNO3)=0.003352mol
m(NaCl)=n(NaCl)*58.5g/mol=0.195975g
p=0.195975g/2.000g=97.9875%
The goal of life is living in agreement with nature.