Question

The osmotic pressure of blood is $7.65atm$ at $37{}^\circ C$. How much glucose should be used per litre for an intravenous injection that is to have the same osmotic pressure as blood?
A. 34.18 g of glucose per litre
B. 44.18 g of glucose per litre
C. 54.18 g of glucose per litre
D. 64.18 g of glucose per litre

Answer 1

To find the weight of glucose that has to be added in 1 litre of the intravenous injection we will first have to determine the number of moles of glucose required. Think about a formula that incorporates osmotic pressure, temperature, and the number of moles of a substance.

Complete step by step solution:
The formula that relates the osmotic pressure with the temperature and the number of moles of a substance is:
$\Pi =CRT$
Where, $\Pi$ is the osmotic pressure, $C$ is the molar concentration, $R$ is the universal gas constant, and $T$ is the ambient temperature. We have been given the values:

& \Pi =7.65atm \\\ & R=0.082L\ atm\text{ }{{K}^{-1}}mo{{l}^{-1}} \\\ & T=37{}^\circ C=310K \\\ \end{aligned}$$ Using these, we will calculate the value of $C$. $$\begin{aligned} & \Pi =CRT \\\ & C=\dfrac{\Pi }{RT} \\\ & C=\dfrac{7.65}{0.082\times 310} \\\ & C=0.301mol/L \\\ \end{aligned}$$ Here, it is given that the volume of the solution is 1 litre, so the molar concentration will be numerically equal to the number of moles of glucose that should be present. n = 0.301 mol/L Now, we have to calculate the amount of glucose to be added. For this, we will use the formula that we use to calculate the number of moles of a given substance. We will require the molecular weight of glucose to calculate the weight of glucose to be added. The molecular formula of glucose is ${{C}_{6}}{{H}_{12}}{{O}_{6}}$. $$\begin{aligned} & \text{Molecular weight of }{{C}_{6}}{{H}_{12}}{{O}_{6}}=(6\times \text{atomic weight of }C)+(12\times \text{atomic weight of }H)+ \\\ & (6\times \text{atomic weight of }O) \\\ & \text{Molecular weight of }{{C}_{6}}{{H}_{12}}{{O}_{6}}=(6\times 12)+(12\times 1)+(6\times 16) \\\ & \text{Molecular weight of }{{C}_{6}}{{H}_{12}}{{O}_{6}}=180 \\\ \end{aligned}$$ We know that, $$\text{no}\text{. of moles = }\dfrac{\text{given weight}}{\text{molecular weight}}$$ So, calculating for the given weight, we get: $$\begin{aligned} & \text{given weight}=\text{no}\text{. of moles}\times \text{molecular weight} \\\ & \text{given weight}=0.301\times 180 \\\ & \text{given weight}=54.18g \\\ \end{aligned}$$ **Hence, the correct answer to this question is C. ‘54.18 g of glucose per litre’** **Additional information:** We can also solve this problem by converting all the units to SI units before carrying out any calculations. Make sure that the value of the universal gas constant is taken according to the units of the other parameters. **Note:** The osmotic pressure is directly proportional to the number of molecules that are present inside the walls of the blood vessels, an estimate of how many molecules are present will be given by the total number of moles of glucose present. The osmotic pressure has to be maintained inside the body for homeostatic equilibrium.