Intravenous, or IV, solutions used in medicine must exert the same osmotic pressure as blood to prevent a net flow of water into or out of the blood cells. The proper concentration for an intravenous NaCl solution is 0.90 g NaCl per 100. mL of solution (sometimes referred to as 0.90% m/v). If the van't Hoff factor of NaCl is ????=1.8, what is the osmotic pressure of blood at body temperature, 37 ∘C?

[tex]\text{Moles of NaCl} = \text{0.90 g} \times \dfrac{\text{1 mol}}{\text{58.44 g}} = 0.0154 \text{ mol}\\\\c = \dfrac{\text{Moles of NaCl}}{\text{Litres of solution}} = \dfrac{0.0154 \text{ mol}}{\text{0.100 L}} = \text{0.154 mol/L}[/tex]

2. Find the osmotic pressure

The formula for osmotic pressure (Π) is

Π = icRT

T = (37 + 273.15) K = 310.15 K

[tex]\Pi = 1.88 \times \text{0.154 mol}\cdot \text{L}^{-1} \times \text{0.083 14 bar}\cdot\text{L}\cdot\text{K}^{-1}\text{mol}^{-1} \times \text{310.15 K} = \text{7.5 bar}\\\\\text{The osmotic pressure of blood at body temperature is }\boxed{\textbf{7.5 bar}}[/tex]

pstnonsonjoku pstnonsonjoku · Answer 2 · 2021-12-14T09:23:46+01:00

The osmotic pressure of the solution is 6.86 atm.

We have the following information;

Concentration of NaCl = 0.90 g NaCl per 100. mL

Van't Hoff factor = 1.8

Temperature = 37 ∘C + 273 = 310 K

The formula for osmotic pressure is = iMRT

Where;

i = Van't Hoff factor

M = molarity

R = gas constant

T = absolute temperature

We have to convert the concentration to mol/L as follows;

Number of moles of NaCl = 0.90 g/58.5 g/mol = 0.015 moles

Volume of the solution = 100 mL or 0.1 L

Concentration in mol/L = 0.015 moles/ 0.1 L = 0.15 mol/L

Substituting into the formula;

π = 1.8 × 0.15 mol/L ×0.082 L atm mol-1 K-1 × 310 K

π = 6.86 atm

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