Đáp án:

$\rm a)$

Lấy $1/10$ ta sẽ có 8 gam rượu nguyên chất và 10 gam nước.

$\rm n_{C_2H_5OH}=\dfrac{8}{46}=\dfrac{4}{23}\ (mol)\\n_{H_2O}=\dfrac{10}{18}=\dfrac{5}{9}\ (mol)$

Phản ứng xảy ra:

$\rm C_2H_5OH+Na\to C_2H_5ONa+\dfrac{1}{2}H_2\\H_2O+Na\to NaOH+\dfrac{1}{2}H_2$

Theo phương trình:

$\rm n_{H_2}=\dfrac{1}{2}×(n_{C_2H_5OH}+n_{H_2O})=\dfrac{151}{414}\ (mol)$

$\rm V_{H_2}=\dfrac{151}{414}×22,4=8,17\ (L)$

$\rm b)$

$\rm V_{C_2H_5OH}=\dfrac{8}{0,8}=10\ (mL)$

$\rm V_{H_2O}=\dfrac{10}{1}=10\ (mL)$

$\rm G^°=\dfrac{10}{10+10}×100=50^°$