Respuesta :
The grams of glucose must be added to 250g of water at 40C to create a solution with a vapor pressure of 54.0 torr is 58.77 gm.
Vapor pressure lowering is a colligative property of solutions. The vapor pressure of a pure solvent is greater than the vapor pressure of a solution containing a non volatile liquid. This lowered vapor pressure leads to boiling point elevation.
The relationship between the relative lowering of vapour pressure and the molecular weight of the solute and solvent is as shown below.
ΔP/P = [tex]\frac{W}{M} X \frac{m}{w}[/tex]
Here M is the molecular weight of glucose which is 180 g/mol.
Here m is the molecular weight of water which is 18 g/mol.
W is mass of glucose
w is mass of water
P is vapor pressure of water
So
(55.3-54)/55.3 = [tex]\frac{W}{180} X \frac{18}{250}[/tex]
(1.3)/55.3 = [tex]\frac{W}{180} X \frac{18}{250}[/tex]
0.023 = [tex]\frac{W}{180} X \frac{18}{250}[/tex]
W = (0.023 x 180 x 250)/18 = 58.77 gm
To learn more about the vapor pressure please click on the link https://brainly.com/question/4463307
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