Question

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?.

Answer 1

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.

What is the number of moles?

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:

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

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

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

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

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

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

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

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