Question

Hòa tan 10,15g tinh thể MgCl2 6H2O vào 120g nước thu đc dung dịch A

Tính C% dung dịch A

Answer 1

\(m_{ddA}=10,15+120=130,15\left(g\right)\)

\(n_{MgCl_2.6H_2O}=\frac{10,15}{203}=0,05\left(mol\right)\)

Ta có: \(m_{MgCl_2}=n_{MgCl_2.6H_2O}=0,05\left(mol\right)\)

\(\Rightarrow m_{MgCl_2}=0,05\times95=4,75\left(g\right)\)

\(\Rightarrow C\%_{MgCl_2}mới=\frac{4,75}{130,15}\times100\%=3,65\%\)

Answer 2

nMgCl2.6H2O = 0.05 mol

=> nMgCl2 = 0.05 mol

=> mMgCl2 = 4.75 g

mdd A = 10.15+ 120 = 130.15 g

C%MgCl2 = 4.75/130.15*100% = 3.65%

Answer 3

m_MgCl2 = \(\frac{10,15}{95+6.18}.95=4,75\)(g)

C% = \(\frac{4,75}{10,15+120}.100\%=\)3,65%

Answer 4

n_{\(MgCl_2.6H_2O\)} = \(\frac{10,15}{203}\) = 0,05 mol

n_{\(MgCl_2.6H_2O\)} = n_{\(MgCl_2\)} = 0,05 mol

m_{\(MgCl_2\)} = 0,05.95 = 4,75 g

m_ddA = 10,15 + 120 = 130,15 g

C%_ddA = \(\frac{4,75}{130,15}.100\%\) = 3,65%