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The solution of glucose commonly used intravenously has a concentration of 5.0% (m/v) glucose. What is the osmolarity of this solution? The molar mass of glucose is 180 g/mol.

The solution of glucose commonly used intravenously has a concentration of 5.0% (m-example-1
User Moulder
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2 Answers

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To find the osmolarity of the 5.0% (m/v) glucose solution, we first need to calculate the molarity of glucose in the solution.

Given:
Concentration of glucose = 5.0% (m/v)
Molar mass of glucose (C6H12O6) = 180 g/mol

The percentage (m/v) represents 5 grams of glucose per 100 mL of solution.

Step 1: Convert the percentage to grams of glucose in 1 liter (1000 mL) of the solution.

5.0% glucose (m/v) = 5 grams of glucose per 100 mL
1 liter (1000 mL) of the solution contains (5 grams/100 mL) * 1000 mL = 50 grams of glucose

Step 2: Calculate the number of moles of glucose in 1 liter (1000 mL) of the solution.

Number of moles of glucose = (mass of glucose) / (molar mass of glucose)
Number of moles of glucose = 50 grams / 180 g/mol ≈ 0.27778 moles

Step 3: Calculate the osmolarity of the solution.

Osmolarity (in osmoles/L or Osm/L) = Number of moles of solute / Volume of solution in liters
Osmolarity = 0.27778 moles / 1 liter = 0.27778 Osm/L or 0.27778 osmol/L

The osmolarity of the 5.0% (m/v) glucose solution is approximately 0.27778 Osm/L or 0.27778 osmol/L.

Hope this helps you.
willkiddhill
User Nevsan
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Answer:

Osmolarity is approximately 0.278 moles/L.

Step-by-step explanation:

Osmolarity is a measure of the concentration of solute particles in a solution. It is defined as the number of moles of solute per liter of solution. Osmolarity is important in biology because it affects the movement of water across cell membranes.

For The Question:

Given:

  • Concentration of glucose solution = 5.0% (m/v) glucose
  • Molar mass of glucose (C6H12O6) = 180 g/mol

Let's calculate the mass of glucose in the solution.

For a 100 mL solution (which is a typical volume used for calculations):

5.0% of 100 mL =
\sf (5)/(100) * 100 = 5\: grams

Converting the mass of glucose to moles.


\boxed{\sf \textsf{Number of moles }= (Mass )/(Molar\: mass)}

Substituting Value,


\sf Number of moles = (5 )/(180) \approx 0.028

Now,

Calculating the osmolarity.


\sf Osmolarity (O) =\frac{ \textsf{Number of moles}} {Volume (in\:\: liters)}

Since the volume is given in milliliters, we need to convert it to liters:


\sf Volume = 100 mL = (100 )/(1000 )= 0.1 liters

Now,


\sf Osmolarity (O) = (0.02778 moles )/(0.1 liters ) \approx 0.28\;\: moles/L

Therefore, the osmolarity of the 5.0% (m/v) glucose solution is approximately 0.278 moles/L.

User Criesto
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