Đáp án:
mctNaCl(2)mctNaCl(2)=331.2100331.2100=6,62 (g)
mctNaCl(tổng)mctNaCl(tổng)=mctNaCl(1)mctNaCl(1)+mctNaCl(2)mctNaCl(2)=29+6,62=35,62 (g)
mddNaCl(sfu)mddNaCl(sfu)=mctNaCl(1)mctNaCl(1)+mddNaCl(tfu)mddNaCl(tfu)=29+331=360 (g)
CphầntrămNaCl(sfu)CphầntrămNaCl(sfu)=35,62.10036035,62.100360≈9,8944%