`a)`

`n_{CuCl_2} = (50)/(135) = (10)/(27) (mol)`

`C_{M(CuCl_2)} = (10)/(27) : 0,45 = (200)/(243) M`

`b)`

`m_{H_2O} = 1.18 = 18(g)`

`V_{dd} = V_{H_2O} = (18)/1 = 18(ml) = 0,018(l)`

`C_{M(H_2SO_4)} = 1/(0,018) = (500)/9 M`

`c)`

`n_{NH_3} = (22,4)/(22,4) = 1(mol)`

`C_{M(NH_3)} = 1/2 = 0,5M`

`d)`

`n_{HCl} = (31,36)/(22,4) = 1,4(mol)`

`C_{M(HCl)} = (1,4)/(0,5) = 2,8M`

Đáp án + Giải thích các bước giải:

a.

$n_{CuCl_2}=\dfrac{50}{135}=\dfrac{10}{27}(mol)$

$C^{M}_{CuCl_2}=\dfrac{\dfrac{10}{27}}{450.10^{-3}}=\dfrac{200}{243}(M)$

b.

$V_{H_2O}=1.1.18=18(ml)$

$C^{M}_{H_2SO_4}=\dfrac{1}{18.10^{-3}}=\dfrac{500}{9}(M)$

c.

$n_{NH_3}=\dfrac{22,4}{22,4}=1(mol)$

$C^{M}_{NH_3}=\dfrac{1}{2}=0,5(M)$

d.

$n_{HCl}=\dfrac{31,36}{22,4}=1,4(mol)$

$C^{M}_{HCl}=\dfrac{1,4}{0,5}=2,8(M)$