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1.6.3: Composición- Conversión de Escalas- Mezclas de Solventes

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    79768
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    Se prepara un disolvente mixto dado (a fijo\(\mathrm{T}\) y\(\mathrm{p}\)) mezclando\(\mathrm{n}_{1} \, \mathrm{V}_{1}^{*}(\ell) \mathrm{m}^{3}\) el líquido 1 y\(\mathrm{n}_{2} \, \mathrm{V}_{2}^{*}(\ell) \mathrm{m}^{3}\) el líquido 2. Supondremos que las propiedades termodinámicas de la mezcla son ideales.

    \[\text { Then volume } \mathrm{V}=\mathrm{n}_{1} \, \mathrm{V}_{1}^{*}(\ell)+\mathrm{n}_{2} \, \mathrm{V}_{2}^{*}(\ell)\]

    Entonces el% en volumen del líquido 2 en la mezcla viene dado por la ecuación (b).

    \[\mathrm{V}_{2} \%=\left[10^{2} \, \mathrm{n}_{2} \, \mathrm{V}_{2}^{*}(\ell)\right] /\left[\mathrm{n}_{1} \, \mathrm{V}_{1}^{*}(\ell)+\mathrm{n}_{2} \, \mathrm{V}_{2}^{*}(\ell)\right]\]

    La masa de un sistema disolvente mixto dado es igual\(\mathrm{w}_{\mathrm{s}}\). Más% en masa del líquido 2 es\(\mathrm{w}_{2}\)%.

    \[\text { Thus } \mathrm{w}_{2} \%=\mathrm{w}_{2} \, 10^{2} /\left(\mathrm{w}_{1}+\mathrm{w}_{2}\right)\]

    \[\text { Mole fraction } \mathrm{x}_{2}=\left(\mathrm{w}_{2} \% / \mathrm{M}_{2}\right) /\left[\frac{\left(10^{2}-\mathrm{w}_{2} \%\right)}{\mathrm{M}_{1}}+\frac{\mathrm{w}_{2} \%}{\mathrm{M}_{2}}\right]\]

    \[\text { Also } \left.\mathrm{V}_{2} \%(\mathrm{mix} ; \text { id })=\left[10^{2} \, \mathrm{w}_{2} / \rho_{2}^{*}(\ell)\right] / \frac{\mathrm{w}_{1}}{\rho_{1}^{*}(\ell)}+\frac{\mathrm{w}_{2}}{\rho_{2}^{*}(\ell)}\right]\]

    Si\(\left(\mathrm{w}_{1}+\mathrm{w}_{2}\right)=100 \mathrm{~kg}\),

    \[\mathrm{V}_{2} \%(\operatorname{mix} ; \mathrm{id})=\left[10^{2} \, \mathrm{w}_{2} \% / \rho_{2}^{*}(\ell)\right] /\left[\frac{\left(10^{2}-\mathrm{w}_{2} \%\right)}{\rho_{1}^{*}(\ell)}+\frac{\mathrm{w}_{2} \%}{\rho_{2}^{*}(\ell)}\right]\]

    Molalidad y fracción molar

    Una mezcla solvente dada tiene masa\(10^{2} \mathrm{~kg}\) se prepara usando\(\mathrm{w}_{2} \mathrm{~kg}\left[=\mathrm{w}_{2} \% \right]\) el líquido 2; nj moles de soluto se disuelven en esta mezcla.

    \[\text { Molality } \mathrm{m}_{\mathrm{j}} / \mathrm{mol} \mathrm{} \mathrm{kg}^{-1}=\mathrm{n}_{\mathrm{j}} / 10^{2}\]

    \[\text { Mole fraction, } \mathrm{x}_{\mathrm{j}}=\mathrm{n}_{\mathrm{j}} /\left\{\left[\left(10^{2}-\mathrm{w}_{2} \%\right) / \mathrm{M}_{1}\right]+\left[\mathrm{w}_{2} \% / \mathrm{M}_{2}\right]+\mathrm{n}_{\mathrm{j}}\right\}\]

    Para soluciones diluídas,\(\mathrm{n}_{\mathrm{j}}<<\left(\mathrm{n}_{1}+\mathrm{n}_{2}\right)\)

    \[\text { Then, } \quad \mathrm{x}_{\mathrm{j}}=\mathrm{n}_{\mathrm{j}} /\left\{\left[\left(10^{2}-\mathrm{w}_{2} \%\right) / \mathrm{M}_{1}\right]+\left[\mathrm{w}_{2} \% / \mathrm{M}_{2}\right]\right\}\]

    \[\text { Or, } \left.\mathrm{x}_{\mathrm{j}}=10^{2} \, \mathrm{m}_{\mathrm{j}} /\left\{\left[10^{2}-\mathrm{w}_{2} \%\right] / \mathrm{M}_{1}\right]+\left[\mathrm{w}_{2} \% / \mathrm{M}_{2}\right]\right\}\]

    Concentración y Molalidad

    Se prepara una solución dada (a fija\(\mathrm{T}\) y\(\mathrm{p}\)) utilizando\(\mathrm{n}_{1}\) moles de líquido 1,\(\mathrm{n}_{2}\) moles de líquido 2 y\(\mathrm{n}_{j}\) moles de un soluto simple (por ejemplo urea) donde\(n_{j}<<\left(n_{1}+n_{2}\right)\).

    \[\text { Mass of mixed solvent } \mathrm{w}_{\mathrm{s}}=\mathrm{n}_{1} \, \mathrm{M}_{1}+\mathrm{n}_{2} \, \mathrm{M}_{2}\]

    \[\text { Mass of system, } w=n_{j} \, M_{j}+n_{1} \, M_{1}+n_{2} \, M_{2}\]

    \[\text { Molality of solute } \mathrm{m}_{\mathrm{j}}=\mathrm{n}_{\mathrm{j}} /\left[\mathrm{n}_{1} \, \mathrm{M}_{1}+\mathrm{n}_{2} \, \mathrm{M}_{2}\right]=\mathrm{n}_{\mathrm{j}} /\left[\mathrm{w}_{1}+\mathrm{w}_{2}\right]\]

    \[\text { Or, } \quad \mathrm{m}_{\mathrm{j}}=\mathrm{n}_{\mathrm{j}} / \mathrm{w}_{\mathrm{s}}\]

    Densidad de la solución\(= \rho\) Masa de la solución\(= \mathrm{w}\)

    \[\text { Volume of solution } \mathrm{V}=\left[\mathrm{n}_{\mathrm{j}} \, \mathrm{M}_{\mathrm{j}}+\mathrm{n}_{1} \, \mathrm{M}_{1}+\mathrm{n}_{2} \, \mathrm{M}_{2}\right] / \rho\]

    \[\text { Concentration of solute } j, \mathrm{c}_{\mathrm{j}}=\mathrm{n}_{\mathrm{j}} \, \rho /\left[\mathrm{n}_{\mathrm{j}} \, \mathrm{M}_{\mathrm{j}}+\mathrm{n}_{1} \, \mathrm{M}_{1}+\mathrm{n}_{2} \, \mathrm{M}_{2}\right]\]

    \[\text { For dilute solutions, } \mathrm{n}_{\mathrm{j}} \, \mathrm{M}_{\mathrm{j}} \ll\left[\mathrm{n}_{1} \, \mathrm{M}_{1}+\mathrm{n}_{2} \, \mathrm{M}_{2}\right]\]

    \[\text { Then, } \mathrm{c}_{\mathrm{j}} \cong \mathrm{n}_{\mathrm{j}} \, \rho /\left[\mathrm{n}_{1} \, \mathrm{M}_{1}+\mathrm{n}_{2} \, \mathrm{M}_{2}\right]\]

    Si la solución se diluye, la densidad de la solución es aprox. igual a la densidad del disolvente\(\rho_{s}\) al mismo\(\mathrm{T}\) y\(\mathrm{p}\).

    \[\text { Hence } \mathrm{c}_{\mathrm{j}} \cong \mathrm{n}_{\mathrm{j}} \, \rho_{\mathrm{s}} /\left[\mathrm{n}_{1} \, \mathrm{M}_{1}+\mathrm{n}_{2} \, \mathrm{M}_{2}\right]\]

    \[\text { Molality of solute } \mathrm{m}_{\mathrm{j}}=\mathrm{n}_{\mathrm{j}} /\left[\mathrm{n}_{1} \, \mathrm{M}_{1}+\mathrm{n}_{2} \, \mathrm{M}_{2}\right]\]

    Entonces\(c_{j} \cong m_{j} \, p_{s}\)


    This page titled 1.6.3: Composición- Conversión de Escalas- Mezclas de Solventes is shared under a Public Domain license and was authored, remixed, and/or curated by Michael J Blandamer & Joao Carlos R Reis via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.