Question

Hòa tan 25g CaCl2 6H2O trong 300ml nước đc dung dịch A có d=1,08g/ml

Nồng độ % dung dịch A

Nồng độ mol đ A

gấp

Answer 1

\(n_{CaCl_2.6H_2O}=\frac{25}{219}\left(mol\right)\)

Ta có: \(n_{CaCl_2}=n_{CaCl_2.6H_2O}=\frac{25}{219}\left(mol\right)\)

\(\Rightarrow m_{CaCl_2}=\frac{25}{219}\times111=12,67\left(g\right)\)

\(m_{H_2O}=300\times1=300\left(g\right)\)

Ta có: \(m_{ddA}=25+300=325\left(g\right)\)

\(C\%_{CaCl_2}mới=\frac{12,67}{325}\times100\%=3,9\%\)

Ta có: \(V_{ddA}=\frac{325}{1,08}=300,93\left(ml\right)=0,30093\left(l\right)\)

\(\Rightarrow C_{M_{CaCl_2}}=\frac{25}{219}\div0,30093=0,38\left(M\right)\)

Answer 2

n_{cacl2.6H2O =} 25/219 (mol) => m_CaCl2 = \(\frac{925}{73}\)(g)

m_A = 300.1,08=324 (g)

=> C% = \(\frac{\frac{925}{73}}{324}.100\%\approx3,9\%\)

C_M = \(\frac{\frac{25}{219}}{0,3}\)=0,381%

Answer 3

n_{\(CaCl_2.6H_2O\)} = \(\frac{25}{219}\) mol

n_{\(CaCl_2\)} = n_{\(CaCl_2.6H_2O\)} = \(\frac{25}{219}\) mol

m_{\(CaCl_2\)} = \(\frac{25}{219}\) . 111 = \(\frac{925}{73}\) g

m_ddA = d.V = 300.1,08 = 324 g

=> C%_ddA = \(\frac{925}{73}\) : 324 . 100% = 3,9%

V_ddA = V_{\(H_2O\)} = 300 ml = 0,3 l

C_M = \(\frac{25}{219}\) : 0,3 = 0,38 M