Question

em 356,4 de água são dissolvidos 68,4g de sacarose c12h22o11. determine as frações molares da sacarose e da água.

quero só o x1 e x2

Answer 1

$Ol\acute{a}~~\mathbb{PERONIKA} \\ \\ Primeiro~sacareos~massa~molar~de~cada~um~deles, veja: \\ \\ \acute{a}gua=H_2O-\ \textgreater \ MM=\underbrace{1.2+16}_{18g/mol} \\ \\ Sacarose=C_{12}H_{22}O_{11}--\ \textgreater \ MM=\underbrace{12.12+1.22+16.11}_{342g/mol} \\ \\ \mathbb{8989898989898989898989898989} \\ Nos~disse~que, em ~356,4g~de~H_2O~sacamos~o~mole~que~corresponde \\ assim: \\ \\ \boxed{ n_1= \frac{m_1}{M_2} }$

$Dados: \\ m_1=356,4g \\ M_1=18g/mol \\ n_1=? \\ Substituindo~dados~na~f\acute{o}rmula~temos: \\ \\ n_1= \frac{356,4g}{18g/mol} \\ \\ \boxed{ n_1=19,8mol} \\ \\ \mathbb{898989898989898989898989898989} \\ Igual~sacaremos~o~mol~que~corresponde ~em~68,4g~de~sacarose, veja: \\ \\ \boxed{n_2= \frac{m_2}{M_2}}$

$Dados: \\ m_2=64,4g \\ M_2=342g/mol \\ n_2=? \\ \\ Substituindo~dados~temos: \\ \\ n_2= \frac{64,4g}{342g/mol} \\ \\ \boxed{n_2=0,2mol} \\ \\ \mathbb{89898989898989898989898989898}\\ agora~calcularemos~''x_1''~~e~~''x_2''$

$Sendo~que: \\ \\\boxed{ x_1= \frac{n_1}{n_1+n_2} }~~~e~~~~\boxed{x_2= \frac{n_2}{n_1+n_2}} \\ \\ subtituindo~n_1~~e~~n_2~temos: \\ \\ x_1= \frac{19,8}{19,8+0,2} ~~~~~~~~~~~~~x_2= \frac{0,2}{19,8+0,2} \\ \\ x_1= \frac{19,8}{20}~~~~~~~~~~~~~~~~~~x_2= \frac{0,2}{20} \\ \\ \underbrace{\boxed{\boxed{x_1=0,99}}~~~~~~~~~~~~~~~~~~\boxed{\boxed{x_2= 0,01}}}_{\underbrace{x_1+x_2=1}_{\boxed{0,99+0,01=1}} }~~~--\ \textgreater \ provando$

$\mathbb{vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv} \\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~Espero~ter~ajudado!!$