Question

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.

Answer 1

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

Answer 2

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.