Câu 6:
$a) n_{NaCl} = 2.0,5=1(mol)$
$m_{NaCl}=58,5.1=58,5(g)$
$b) n_{KNO3} = 0,5.3=1,5(mol)$
$m_{KNO3} = 101.1,5=151,5(g)$
$c) m_{ctFeCl2} = \dfrac{50.8}{100}=4(g)$
$n_{FeCl2}=\dfrac{4}{27}(mol)$
$d) m_{ctHCl} = \dfrac{200.12}{100} =24(g)$
$n_{HCl} = \dfrac{24}{36,5}=\dfrac{48}{73}(mol)$
Câu 7:
$a) C_{M_{KCl}}=\dfrac{1}{0,75}=1,33(M)$
$b) C_{M_{MgCl_2}}=\dfrac{0,5}{1,5}=0,33(M)$
$c) C_{CuSO_4}=\dfrac{400}{160}=2,5(mol) ⇒ C_{M_{CuSO_4}}=\dfrac{2,5}{4}=0,625(M)$
$d) C_{M_{Na_2CO_3}}=\dfrac{0,06}{1,5}=0,04(M)$
Câu 8:
$a) C\%NaCl = \frac{mNaCl}{mdd} .100\% = \frac{10}{500} .100\% = 2\%$
$b) C\%NaNO3 = \frac{mNaNO3}{mdd} .100\% = \frac{45}{1,5.1000} .100\% = 3\%$
$c) C\%FeSO4 = \frac{mFeSO4}{mdd} .100\% = \frac{85}{1200} .100\% = 7,08\%$
d) $mZnCl2 = 0,1.136 = 13,6 gam$
$⇒ C\%ZnCl2 = \frac{mZnCl2}{mdd} .100\% = \frac{13,6}{300} .100\% = 4,53\% $