Đáp án:
\( {d_{X/{O_2}}} = 1,494\)
Giải thích các bước giải:
Ta có:
\({m_X} = {m_{C{O_2}}} + {m_{S{O_2}}} + {m_{{N_2}O}} = 50,2{\text{ gam}}\)
Ta có:
\({n_{C{O_2}}} = \frac{{22}}{{44}} = 0,5{\text{ mol;}}{{\text{n}}_{S{O_2}}} = \frac{{12,8}}{{64}} = 0,2{\text{ mol;}}{{\text{n}}_{{N_2}O}} = \frac{{15,4}}{{44}} = 0,35{\text{ mol}}\)
\({n_X} = {n_{C{O_2}}} + {n_{S{O_2}}} + {n_{{N_2}O}} = 0,5 + 0,2 + 0,35 = 1,05{\text{ mol}}\)
\( \to {M_X} = \frac{{{m_X}}}{{{m_X}}} = \frac{{50,2}}{{1,05}} = 47,81\)
\( \to {d_{X/{O_2}}} = \frac{{{M_X}}}{{{M_{{O_2}}}}} = \frac{{47,81}}{{32}} = 1,494\)
