Respuesta :

Dejanras
m(dextrose) = 50 g.
V(solution) = 1 L.
n(dextrose) = 50 g ÷ 180 g/mol.
n(dextrose) = 0,27 mol.
Osmotic concentration (osmolarity) is a measure of how many osmoles of particles of solute it contains per liter.
The osmolarity = n(dextrose) ÷ V(solution).
The osmolarity = 0,27 mol ÷ 1 L.
The osmolarity = 0,27 mol/L · 1000 mmol/m.
The osmolarity (dextrose) = 270 mosm/L.
The osmolarity (dextrose monohydrate) = 50 g÷197 g/mol·1000 =254mosm/L
superman1987
The answer is 252 mOsm/L.
when Osmolarity is (the osmotic concentration) which measure of solute concentration as it is the number of (OSM) of solute per L of the solution. 

According to the equation of osmolarity for nonelectrolytes (dextrose: glucose monohydrate):
Osmolarity = (g/L) / MW x 1000
when the MW of glucose monohydrate = 198.17 so,
by  substitute:
 ∴ Osmolarity = 50 (g/L) / 198.17 x  1000 = 252 mOsm/L


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