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lozanna [386]
3 years ago
6

Si desea preparar 300 gramos de una solución de azúcar en agua que tenga una concentración de 20% m/m en cuanto gramos de soluto

de azúcar se necesita?
Chemistry
1 answer:
Digiron [165]3 years ago
5 0

Answer:

m_{soluto}=60g

Explanation:

¡Hola!

En este caso, al considerar la unidad de concentración de porcentaje masa/masa, podemos escribir su fórmula como:

\%m/m=\frac{m_{soluto}}{m_{solucion}}*100\%

Podemos identificar la masa de soluto (azúcar) como la incógnita, y resolverla como se muestra a continuación:

m_{soluto}=\frac{\%m/m*m_{solucion}}{100\%}\\\\m_{soluto}=\frac{20\%*300g}{100\%}\\\\m_{soluto}=60g

¡Saludos!

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Formula used to calculate the final temperature is as follows.

\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}

Substitute the values into above formula as follows.

\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}\\\frac{2.7 atm \times 3.1 L}{300 K} = \frac{P_{2} \times 9.4 L}{610 K}\\P_{2} = \frac{5105.7}{2820} atm\\= 1.81 atm

Now, moles present upon heating the cylinder are as follows.

P_{2}V_{2} = n_{2}RT_{2}\\1.81 atm \times 9.4 L = n_{2} \times 0.0821 L atm/mol K \times 610 K\\n_{2} = \frac{17.014}{50.081} mol\\= 0.34 mol

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