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

Answer:

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:

$P_S = P^0_A \times X_A \ \ \ OR \ \ \ P_A = P^0_A \times X_A$

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