Đáp án + Giải thích các bước giải:

Câu 4:

$n_{NaCl}=\frac{m}{M}=\frac{52,65}{58,5}=0,9(mol)$

$\to V_{dd\,NaCl}=\frac{n}{C_M}=\frac{0,9}{3}=0,3(l)=300(ml)$

Câu 5:

$a\big)$

$n_{Mg}=\frac{18}{24}=0,75(mol)$

$Mg+H_2SO_4\to MgSO_4+H_2$

Theo PT: $n_{H_2SO_4}=n_{Mg}=0,75(mol)$

Ta có $m_{dd\,H_2SO_4}=V_{dd}.D=75.1,2=90(g)$

$\to C\%_{H_2SO_4}=\frac{0,75.98}{90}.100\%\approx 81,67\%$

$b\big)$

$m_{dd\,H_2SO_4}=85.1,2=102(g)$

$\to C\%_{H_2SO_4}=\frac{0,75.98}{102}.100\%\approx 72,06\%$

Đáp án `+` Giải thích các bước giải:

Câu 4:

$n_{NaCl}=\dfrac{m}{M}=\dfrac{52,65}{58,5}=0,9(mol)$

$\to V_{dd\,NaCl}=\dfrac{n_{NaCl}}{C_{M_{NaCl}}}=\dfrac{0,9}{3}=0,3(l)=300(ml)$

Câu 5:

$a)$ $n_{Mg}=\dfrac{18}{24}=0,75(mol)$

$Mg+H_2SO_4\to MgSO_4+H_2$

Theo PT: $n_{H_2SO_4}=n_{Mg}=0,75(mol)$

`=>` $m_{dd\,H_2SO_4}=V_{dd}.D=75.1,2=90(g)$

`=>` $C\%_{H_2SO_4}=\dfrac{0,75.98}{90}.100\%\approx 81,67\%$

$b)$ $m_{dd\,H_2SO_4}=85.1,2=102(g)$

`=>` $C\%_{H_2SO_4}=\dfrac{0,75.98}{102}.100\%\approx 72,06\%$