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OleMash [197]
2 years ago
13

How many grams of sucrose, c12h22o11, must be added to 500. G of water at 100°c to change the vapor pressure to 752 mm hg?.

SAT
1 answer:
denpristay [2]2 years ago
8 0

The mass number of sucrose that must be added to 500 g of water is 100.979 grams

To be able to solve this question, we need to use the number of moles concept and mole fraction concept.

<h3>What is the number of moles?</h3>

The number of moles is related to the mass of the substance divided by the molar mass of the substance.

From the given parameters;

The number of moles of water = mass/molar mass

  • Mass of water = 500 g
  • the molar mass of water = 18.02 g/mol

number of moles = 500 g/ 18.02 g/mol

number of moles = 27.75 moles

However,

  • the standard vapor pressure of the solvent \mathbf{P^0_{water}} = 760 mmHg
  • the vapor pressure of the solute = 752 mmHg

Using the mole concept:

\mathbf {X_{C_{12}H_{22}O_{11}} = \dfrac{n_{C_{12}H_{22}O_{11}}}{n_{H_2O}+n_{C_{12}H_{22}O_{11}}} }

By relative lowering of vapor pressure:

\mathbf{\dfrac{P^0_{H_2O} - P_{H_2O}}{P^0_{H_2O}} = X_{C_{12}H_{22}O_{11}}}

\mathbf{\dfrac{760 -752}{760} = \dfrac{n_{C_{12}H_{22}O_{11}}}{27.75 +n_{C_{12}H_{22}O_{11}}} }}

\mathbf{0.01053 = \dfrac{n_{C_{12}H_{22}O_{11}}}{27.75 +n_{C_{12}H_{22}O_{11}}} }}

\\ \\ \mathbf{0.01053 (27.75) +0.01053 (n_{C_{12}H_{22}O_{11}} ) = n_{C_{12}H_{22}O_{11}}}

\\ \\ \mathbf{0.2922075 =0.98947&#10; (n_{C_{12}H_{22}O_{11}} )}

\\ \\ \mathbf{ (n_{C_{12}H_{22}O_{11}} ) = \dfrac{0.2922075 }{0.98947}}

\\ \\ \mathbf{ (n_{C_{12}H_{22}O_{11}} ) =0.295 \ moles}

The mass of sucrose = number of moles × molar mass

The mass of sucrose = 0.295 moles × 342.3 g/mol

The mass of sucrose = 100.979 grams

Learn more about number of moles here:

brainly.com/question/13314627

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