nH2SO4 = 4.9/98=0.05 mol
CM H2SO4 = 0.05/1 = 0.05M
mH2O = 1000 g
mdd = 4.9 + 1000 = 1004.9 g
C%H2SO4 = 4.9/1004.9*100% = 0.48%
\(n_{H_2SO_4}=\frac{4,9}{98}=0,05\left(mol\right)\)
\(C_{M_{H_2SO_4}}=\frac{0,05}{1}=0,05\left(M\right)\)
\(m_{H_2O}=0,001\times1000=1\left(kg\right)=1000\left(g\right)\)
\(m_{dd}saupư=4,9+1000=1004,9\left(g\right)\)
\(C\%_{H_2SO_4}=\frac{4,9}{1004,9}\times100\%=0,488\%\)
Có 1l nước= 1kg nước=1000g
K/l dd H2SO4:
.................m(ddH2SO4)=1000+4,9=1004,9(g)
C%=4,9/1004,9.100\(\approx0,49\left(\%\right)\)
Số mol: nH2SO4=4,9/98=0,02(mol)
CM=0,02/1=0,02(M)
\(n_{H_2SO_4}=\frac{4,9}{98}=0,05\left(mol\right)\\ \rightarrow C_M=\frac{0,05}{1}=0,05\left(M\right)\)
\(C\%=\frac{4,9}{1.1000+4,9}.100\%=0,48\left(\%\right)\)
n\(H_2SO_4\) = \(\frac{4,9}{98}\) = 0,05 mol
CM\(H_2SO_4\) = \(\frac{0,05}{1}=0,05M\)
m\(H_2O\) = 1000 g
mdd sau pư = 4,9 + 1000 = 1004,9 g
C% = \(\frac{49}{1004,9}.100\%\) = 0,48 %