Đáp án:

$C\%=\dfrac{m_{CH_3COOH}}{V_{dd}.D_{dd}}.100\%=\dfrac{80,8}{1000.1,0097}.100\%=8\%$ 

$C_M=\dfrac{m_{CH_3COOH}}{M_{CH_3COOH}.V_{dd}}=\dfrac{80,8}{60.1}=1,35M$

$C_m=\dfrac{m_{CH_3COOH}}{M_{CH_3COOH}.m_{H_2O}}=\dfrac{m_{CH_3COOH}}{M_{CH_3COOH}.(D_{dd}.V_{dd}-80,8).10^{-3}}=\dfrac{80,8}{60.(1000.1.0097-80,8).10^{-3}}=1,45 mol.kg^{-1}$

$C\%=\dfrac{80.8}{1000 \times 1.0097} \times 100\%=8\%$

$C_M=\dfrac{10dC\%}{M}=\dfrac{10 \times 1.0097 \times 8}{60}=1.346(M)$

$C_m=\dfrac{\dfrac{80.8}{60}}{1000 \times 1.0097 \times 91\%} \times 1000=1.466(mol/kg)$