Đáp án:

\(\begin{array}{l}
1)\\
a)\\
\bar M = 192,22\\
b)\\
\bar M = 79,9862\\
c)\\
\bar M = 39,1372\\
d)\\
{V_{Ar}} = 11,20224l\\
2)\\
\bar M = 79,92
\end{array}\)

3)

\(\bar M = 20,18\)

Giải thích các bước giải:

\(\begin{array}{l}
1)\\
a)\\
\bar M = \dfrac{{191 \times 39 + 193 \times 61}}{{100}} = 192,22\\
b)\\
\bar M = \dfrac{{79 \times 50,69 + 81 \times 49,31}}{{100}} = 79,9862\\
c)\\
\bar M = \dfrac{{93,08 \times 39 + 0,12 \times 40 + 41 \times 6,8}}{{100}} = 39,1372\\
d)\\
\bar M = \dfrac{{36 \times 0,337 + 38 \times 0,063 + 40 \times 99,6}}{{100}} = 39,98526\\
{n_{Ar}} = \dfrac{{20}}{{39,98526}} = 0,5001mol\\
{V_{Ar}} = 0,5001 \times 22,4 = 11,20224l\\
2)\\
{A_1} = 35 + 44 = 79\\
{A_2} = 35 + 44 + 2 = 81\\
\% {A_1} = \dfrac{{27}}{{50}} \times 100\%  = 54\% \\
\% {A_2} = 100 - 54 = 46\% \\
\bar M = \dfrac{{79 \times 54 + 81 \times 46}}{{100}} = 79,92
\end{array}\)

\(\begin{array}{l}
3)\\
\text{Tổng số hạt của X là 32}\\
2{p_X} + {n_X} = 32 \Rightarrow {n_X} = 32 - 2{p_X}\\
1 \le \dfrac{{{n_X}}}{{{p_X}}} \le 1,5 \Rightarrow 1 \le \dfrac{{32 - 2{p_X}}}{{{p_X}}} \le 1,5\\
 \Rightarrow 9,14 \le {p_X} \le 10,6\\
 \Rightarrow {p_X} = 10\\
{n_X} = 32 - 10 \times 2 = 12\\
{A_X} = 10 + 12 = 22\\
{n_Y} = 12 - 2 = 10\\
{A_Y} = 10 + 10 = 20\\
\% X = \dfrac{{18}}{{18 + 182}} \times 100\%  = 9\% \\
\% Y = 100 - 9 = 91\% \\
\bar M = \dfrac{{22 \times 9 + 20 \times 91}}{{100}} = 20,18
\end{array}\)