Question

. What will be the osmotic pressure of a solution that was prepared at 91°F by dissolving 534 grams of aluminum hydroxide in enough water to generate 2.784 ml of solution.

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Answer to a math question . What will be the osmotic pressure of a solution that was prepared at 91°F by dissolving 534 grams of aluminum hydroxide in enough water to generate 2.784 ml of solution.

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Esmeralda
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Para determinar la presión osmótica de esta solución, primero necesitaremos calcular la concentración de la solución en términos de moles por litro. Luego, utilizaremos la fórmula de la presión osmótica para obtener el resultado final.

Paso 1: Calcular los moles de hidróxido de aluminio (Al(OH)3)
Para esto, necesitamos convertir los gramos de Al(OH)3 a moles. Utilizaremos la masa molar del compuesto, que es de 78 g/mol.

\text{Moles de Al(OH)3} = \frac{{534 \, \text{g}}}{{78 \, \text{g/mol}}} = 6.846 \, \text{mol}

Paso 2: Calcular el volumen en litros
Dado que el volumen de la solución se da en mililitros, necesitamos convertirlo a litros. Utilizaremos la relación de conversión 1L = 1000 ml.

\text{Volumen (litros)} = \frac{{2784 \, \text{ml}}}{{1000 \, \text{ml/L}}} = 2.784 \, \text{L}

Paso 3: Calcular la concentración en moles por litro
Utilizando los moles de Al(OH)3 y el volumen en litros, podemos calcular la concentración de la solución.

\text{Concentración (M)} = \frac{{6.846 \, \text{mol}}}{{2.784 \, \text{L}}} = 2.459 \, \text{M}

Paso 4: Calcular la presión osmótica
La fórmula para calcular la presión osmótica es:

\text{Presión osmótica} = \text{Concentración (M)} \times \text{R} \times \text{Temperatura (K)}

Donde R es la constante de los gases ideales, que tiene un valor de 0.0821 L·atm/(mol·K). Necesitamos primero convertir la temperatura a kelvin. Utilizando la fórmula:

\text{Temperatura (K)} = \text{Temperatura (°C)} + 273.15

\text{Temperatura (K)} = 91°C + 273.15 = 364.15 \, \text{K}

Sustituyendo los valores conocidos en la fórmula de la presión osmótica:

\text{Presión osmótica} = 2.459 \, \text{M} \times 0.0821 \, \text{L·atm/(mol·K)} \times 364.15 \, \text{K}

\text{Presi^^f3n osm^^f3tica}=73.52\,\text{atm}

Respuesta: La presión osmótica de la solución es de 73.52 atm.

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