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a/ \(n_{NaOH}=2V\)
\(\Rightarrow m_{NaOH}=2V.40=80V\)
\(m_{dd}=1000V.1,44=1440\)
\(\Rightarrow C\%=\frac{80V}{1440V}=5,56\%\)
b/ \(n_{H_2SO_4}=8V\)
\(\Rightarrow m_{H_2SO_4}=8V.98=784V\)
\(m_{dd}=1000V.1,44=1440V\)
\(\Rightarrow C\%=\frac{784V}{1440V}=54,44\%\)
c/\(n_{CaCl_2}=2,487V\)
\(\Rightarrow m_{CaCl_2}=2,487V.111=276,057V\)
\(m_{dd}=1000V.1,2=1200V\)
\(\Rightarrow C\%=\frac{276,057V}{1200V}=23\%\)
a.\(n_{NaOH}=\dfrac{8}{40}=0,2mol\)
\(V_{dd}=\dfrac{120}{1,2}=100ml=0,1l\)
\(C_{M_{NaOH}}=\dfrac{0,2}{0,1}=2M\)
b.\(n_{NaOH}=\dfrac{21,6}{40}=0,54mol\)
\(V_{dd}=\dfrac{180}{1,2}=150ml=0,15l\)
\(C_{M_{NaOH}}=\dfrac{0,54}{0,15}=3,6M\)
PTHH: H2SO4+2NaOH→Na2SO4+2H2O
nH2SO4=0,25×1=0,25 mol.
Theo pt: nNaOH=2nH2SO4=0,5 mol.
Theo pt: nNa2SO4=nH2SO4=0,25 mol.
Vdd spư=250+250=500 ml=0,5 lít.
⇒CM NaOH=0,5/0,25=2 M.
CM Na2SO4=0,25/0,5=0,5 M.
\(a,C_{M\left(NaOH\right)}=\dfrac{0,3}{0,5}=0,6M\\ b,n_{NaOH}=\dfrac{24}{40}=0,6\left(mol\right)\\ C_{M\left(NaOH\right)}=\dfrac{0,6}{0,4}=1,5M\)
\(a,m_{KOH}=\dfrac{28.10}{100}=2,8\left(g\right)\\ \rightarrow n_{KOH}=\dfrac{2,8}{56}=0,05\left(mol\right)\\ b,C\%=\dfrac{36}{144+36}.100\%=20\%\\ c, n_{NaOH}=\dfrac{0,8}{40}=0,02\left(mol\right)\\ \rightarrow C_{M\left(NaOH\right)}=\dfrac{0,02}{0,08}=0,25M\)
\(a,m_{KOH}=\dfrac{28.10}{100}=2,8\left(g\right)\\ n_{KOH}=\dfrac{2,8}{56}=0,05\left(mol\right)\\ C\%=\dfrac{36}{36+144}.100\%=20\%\\ C_M=\dfrac{0,8}{0,08}=10M\)
\(m_{NaOH}=\frac{120.15\%}{100}=18\left(g\right)\) => \(n_{NaOH}=\frac{18}{40}=0,45\left(mol\right)\)
\(V_{dd}=\frac{120}{1,2}=100\left(ml\right)=0,1\left(l\right)\)
=> \(C_M=\frac{0,45}{0,1}=4,5M\)