D = mdd/V ---> mdd = D.V = 1,28.200 = 256 gam. ---> mCaCl2 = mdd.C%/100 = 256.30/100 = 76,8gam.

- Thanks bn nha

a) 300 ml nước ~ 300g nước 
m dd = 25+300 = 325(g) 
n CaCl2.6H2O = n CaCl2 =  25/219 = 0.11 (mol) 
m CaCl2 = 0.11*111 = 12.21(g) 
C%dd = 12.21/325*100% = 3.76% 
V dd = 325/1.08 = 300.93(ml) = 0.3(l) 
CM = 0.11/0.3 = 0.37M 

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

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

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

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%

Nồng độ \(CaCl_2\) trong \(CaCl_2.6H_2O\) là:

\(C\%=\dfrac{M_{CaCl_2}}{M_{CaCl_2.6H_2O}}\cdot100\%=\dfrac{111}{219}\cdot100\%=50,68\%\)

Sơ đồ chéo:

\(CaCl_2.6H_2O\)    \(m_1\)    50,68                   40

                                          40

\(H_2O\)                 \(m_2\)    0                          10,68

\(\Rightarrow\dfrac{m_1}{m_2}=\dfrac{40}{10,68}\)  (*)

\(m_{dd}=V\cdot D=10\cdot1,395=13,95g=m_1+m_2\)

\(\Rightarrow m_2=13,95-m_1\) Thay vào (*) ta được:

\(\Rightarrow\dfrac{m_1}{13,95-m_1}=\dfrac{40}{10,68}\Rightarrow m_1=11,01g\)

\(\Rightarrow m_2=13,95-11,01=2,94g\Rightarrow n_{H_2O}=0,163mol\)

\(V_{H_2O}=0,163\cdot22,4=3,65l\)

\(m_{dd} = 25 + 300 = 325(gam)\\ n_{CaCl_2} = \dfrac{325.3,9\%}{111} = \dfrac{169}{1480}\\ \Rightarrow n_{CaCl_2.nH_2O} = n_{CaCl_2} = \dfrac{169}{1480}(mol)\\ \Rightarrow (111 + 18n).\dfrac{169}{1480} = 25\\ \Rightarrow n = 6\)

CTPT tinh thể : \(CaCl_2.6H_2O\)

CTPT tinh thể : 

$n_{Na_2CO_3} = n_{Na_2CO_3.10H_2O} = \dfrac{28,6}{286} = 0,1(mol)$

$C_{M_{Na_2CO_3}} = \dfrac{0,1}{0,2} = 0,5M$
$m_{dd} = D.V = 200.1,05 = 210(gam)$

$C\%_{Na_2CO_3} = \dfrac{0,1.106}{210}.100\% = 5,05\%$

\(m_{dd}\)=1,05.200=210 g

=>C%dd =\(\dfrac{28,6}{210}\) .100% =13,62%

Mặt khác :  200ml=0,2l

M_ct=23.2+12+16.3+10.(1.2+16)=286      (M nguyên tử khối )

=>n_ct=\(\dfrac{28,6}{286}\) =0,1 mol

=>C_M=\(\dfrac{0,1}{0,2}\) =0,5M