Question

ayudenme en la 3 por favor!!!

Answer 1

Fórmula: C12H22O11 (SACAROSA)

Masa molar: 342,29 g/mol

Masa molar agua: 18 g/mol

masa sacarosa = 171.2 g (soluto)

masa agua = 1000 g (solvente)

P v del agua = 17.535 mmHg / 760 = 0.023 atm.

fórmula: P° solc. = P° solvente ·  X solvente

1. calcular moles

n = masa/Mm

n soluto = 171.2 g / 342.29 g/mol = 0.500 mol

n solvente = 1000 g / 18 g/mol = 55.556 mol

2. calcular moles totales

moles totales = moles soluto + moles solvente

moles totales = 0.500 mol + 55.556 mol

moles totales = 56.056

3. calcular fracción molar del solvente:  Xsvte: nsvte / n totales

X svte = 55.556 moles / 56.056 moles

X svte = 0.99

NOTA: LA Mm de la sacarosa de 342.23 g/mol

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resolviendo el ejercicio con la Mm 343.3 g/mol

Fórmula: C12H22O11 (SACAROSA)

Masa molar: 343.3 g/mol

Masa molar agua: 18 g/mol

masa sacarosa = 171.2 g (soluto)

masa agua = 1000 g (solvente)

P v del agua = 17.535 mmHg / 760 = 0.023 atm.

fórmula: P° solc. = P° solvente • X solvente

1. calcular moles

n = masa/Mm

n soluto = 171.2 g / 343.3 g/mol = 0.498 mol

n solvente = 1000 g / 18 g/mol = 55.556 mol

2. calcular moles totales

moles totales = moles soluto + moles solvente

moles totales = 0.500 mol + 55.556 mol

moles totales = 56.054

3. calcular fracción molar del solvente: X svte: n svte / n totales

X svte = 55.556 moles / 56.054 moles

X svte = 0.99

4. calcular la P° de la solución

P° solc. = 0.023 atm x 0.99

P° solc. = 0.023 atm.

4. calcular la P° de la solución

P° solc. = 0.023 atm x 0.99

P° solc. = 0.023 atm.