Question

Suppose 20.23 g of glucose are dissolved in 95.75 g of water at 27.0 OC. Glucose is nonvolatile (has no vapor pressure) and has a molecular wt of 180.2 g/mole. Water has a pure vapor pressure (PA O ) of 26.7 mm Hg at 27.0 OC and a molecular weight of 18.02 g/mole. Find the moles of each component. Then, determine the mole fractions (XGlucose and Xwater) of each component using XA What should XGlucose + Xwater equal? Finally, use Raoult’s Law to determine the total vapor pressure of the solution: PA = (PA O )(XA) where A is H2O.

Answer 1

Step-by-step explanation:

From the information given :

we can understand the solute is glucose and the solvent is water,

So, the weight of glucose = 20.23 g

the molecular weight of glucose = 180.2 g/mol

weight of water = 95. 75 g

the molecular weight of water = 18.02 g/mol

pure vapor pressure of water
`P_A = 26.7 \ mmHg` at 27°C

moles of glucose = weight of glucose/ molecular weight of glucose

= 20.23/180.2

= 0.11 mole

moles of water = weight of water / molecular weight of water

= 95.75/18.02

= 5.31 mole

mole fraction of glucose
`X_(glucose) =` (moles of glucose)/(moles of glucose+ moles of water)

`X_(glucose) =` 0.11/(0.11 + 5.31)

`X_(glucose) =` 0.0203

mole fraction of glucose
`X_(water) =` (moles of water)/(moles of water+ moles of glucose)

`X_(water) =` 5.31/ (5.31 + 0.11)

`X_(water) =` 0.9797

Using Raoult's Law:

`P_S = P^0_A * X_A \ \ \ OR \ \ \ P_A = P^0_A * X_A`

where:

`P_S` = vapor pressure of the solution

`P_A` = total vapor pressure of the solution

`P^0_A`= vapor pressure of the solvent in the pure state

`X_A` = mole fraction of solvent i.e. water

`P_A =` 95.75 × 0.9797

`P_A =` 93.81 mmHg

the total vapor pressure of the solution = 93.81 mmHg