1.
Bài ra: \(m_{CuSO_4.5H_2O}=25g\)
Có: \(M_{C\text{uS}O_4.5H_2O}=160+5.18=250\left(\dfrac{g}{mol}\right)\)
\(\Rightarrow n_{CuSO_4.5H_2O\:}=\dfrac{25}{250}=0,1\left(mol\right)\)
\(\Rightarrow n_{C\text{uS}O_4}=1.0,1=0,1\left(mol\right)\)
\(\Rightarrow m_{C\text{uS}O_4}=0,1.160=16\left(g\right)\)
\(\Rightarrow m_{H_2O}=25-16=9\left(g\right)\)
Vậy \(m_{ct}=m_{C\text{uS}O_4trong.tinh.th\text{ể}}=16\left(g\right)\)
\(m_{dm}=100+9=109\left(g\right)\)
\(m_{\text{dd}}=100+25=125\left(g\right)\)
2.(bạn ghi sai rồi Ba(OH)2chứ ko có (BaOH)2nhé)
Có: \(m_{Ba\left(OH\right)_2}=34,2\left(g\right)\)
\(m_{H_2O}=200g\)
\(\Rightarrow m_{\text{dd}}=34,2+200=234,2\left(g\right)\)
\(\Rightarrow C\%_{_{ }}=\dfrac{34,2}{234,2}.100\%\approx14,603\%\)
