1) \(C\%_{ddKOH}=\dfrac{0,4.56}{0,4.56+200}.100\%=10,07\%\)
2) \(C\%_{ddFe2\left(SO4\right)3}=\dfrac{1.400}{400+500}.100\%=44,4\%\)
3) \(m_{NaCl}=\dfrac{0,9.500}{100}=4,5\left(g\right)\)
4) \(m_{ddNaOH}=\dfrac{4.100}{20}=20\left(g\right)\)
\(\Rightarrow m_{H2O}=20-4=16\left(g\right)\)
5) \(n_{CuSO4}=\dfrac{16}{160}=0,1\left(mol\right)\)
\(\Rightarrow C_{MCuSO4}=\dfrac{0,1}{0,5}=0,2\left(M\right)\)
6) \(n_{KOH}=1.0,2=0,2\left(mol\right)\Rightarrow m_{KOH}=0,2.56=11,2\left(g\right)\)
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7) 500g NaCl 20% \(\Rightarrow m_{NaCl}=20\%.500=100\left(g\right)\)
250g NaCl 10% \(\Rightarrow m_{NaCl}=10\%.250=25\left(g\right)\)
\(\Rightarrow m_{\left(dung.dịch.sau.khi.trộn\right)}=\dfrac{25+100}{300+250}.100=22,73\%\)
8) 300g KOH 20% \(\Rightarrow m_{KOH}=20\%.300=60\left(g\right)\)
250g KOH 10% \(\Rightarrow m_{KOH}=10\%.250=25\left(g\right)\)
\(\Rightarrow m_{\left(dung.dịch.sau.khi.trộn\right)}=\dfrac{60+25}{300+250}.100=15,45\%\)
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\(9.n_{NaCl\left(sau\right)}=0,1.1+0,2.2=0,5mol\\ V_{NaCl\left(sau\right)}=0,1+0,2=0,3l\\ C_{M_{NaCl\left(sau\right)}}=\dfrac{0,5}{0,3}=\dfrac{5}{3}M\)
\(10.n_{HCl\left(sau\right)}=0,1.1+0,4.2=0,9mol\\ V_{HCl\left(sau\right)}=0,1+0,4=0,5l\\ C_{M_{HCl\left(sau\right)}}=\dfrac{0,9}{0,5}=1,8M\)
