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viktelen [127]
3 years ago
11

The osmolarity of a 1-l bottle of 5% dextrose solution is _____ mosm/l.

Chemistry
2 answers:
Pachacha [2.7K]3 years ago
7 0
M(dextrose) = 50 g.
V(solution) = 1 L.
n(dextrose) = 50 g ÷ 180 g/mol.
n(dextrose) = 0,27 mol.
Osmotic concentration (osmolarity)<span> is a measure of how many </span><span>osmoles of particles of solute</span><span> it contains </span>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
creativ13 [48]3 years ago
3 0
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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Explanation:

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E_a=-\ln \frac{1.5\times \:10^3}{4.0\times \:10^3}\:\times \frac{8.314}{\left(\frac{1}{278.15}-\frac{1}{298.15}\right)}

E_a=-\frac{8.314\ln \left(\frac{1.5\times \:10^3}{4\times \:10^3}\right)}{\frac{1}{278.15}-\frac{1}{298.15}}

E_a=-\frac{689483.53266 \ln \left(\frac{1.5}{4}\right)}{20}

E_a=33813.28\ J/mol

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