Question

The $ {D_5}W $ refers to one of the solutions used as an intravenous fluid. It is a 5% by mass solution of dextrose, $ {C_6}{H_{12}}{O_6} $ ​ in water. The density of $ {D_5}W $ is $ 1.08{\text{ mg/L}} $ . The molarity of the solution is:
(A) $ 0.3M $
(B) $ 0.6M $
(C) $ 0.28M $
(D) $ 0.26M $

Answer 1

Hint: Calculate the number of moles of dextrose by using molecular weight and mass percentage values. Volume is calculated from the density of the fluid. These values are then computed, to find the molarity.

Formula Used: $ {\text{Molarity = }}\dfrac{{{\text{No}}{\text{. of moles of solute}}}}{{{\text{Volume of solution}}}} $

Complete step by step solution:
Molarity of a given solution is defined as the total number of moles of solute per litre of solution. The molality of a solution is dependent on the changes in physical properties of the system such as pressure and temperature as unlike mass, the volume of the system changes with the change in physical conditions of the system.
Thus, $ {\text{Molarity = }}\dfrac{{{\text{No}}{\text{. of moles of solute}}}}{{{\text{Volume of solution}}}} $ .
It has been given that, $ {D_5}W $ is a 5% by mass solution of dextrose, $ {C_6}{H_{12}}{O_6} $ ​ in water. The density of $ {D_5}W $ is $ 1.08{\text{ mg/L}} $ , 5% by mass implies 5g of the solution in 1000g of water.
Molecular mass of dextrose $ {C_6}{H_{12}}{O_6} $ is, $ \left( {6 \times 12} \right) + \left( {1 \times 12} \right) + \left( {6 \times 16} \right) = 180{\text{g}} $ where the atomic mass of carbon $ C = 12{\text{g}} $ , hydrogen $ H = 1{\text{g}} $ and oxygen $ = 16{\text{g}} $ .
The number of moles of $ {C_6}{H_{12}}{O_6} $ can be calculated from here.
 $ \therefore $ The number of moles of $ {C_6}{H_{12}}{O_6} = \dfrac{5}{{180}} = 0.0278 $ .
The density of the solution is $ \rho = 1.08{\text{ mg/L}} $ .
We know that the density of a solution is its mass per unit volume. Thus, we can write,
 $ \rho = \dfrac{m}{V} $ where $ m $ is the mass of the solution per unit volume $ V $ .
Thus, volume $ V = \dfrac{m}{\rho } = \dfrac{{100}}{{1.08}} = 92.525{\text{ mL}} $ .
From here, we can calculate the molarity. We know that,
 $ {\text{No}}{\text{. of moles of solute = 0}}{\text{.0278}} $ and $ {\text{Volume of solution = 92}}{\text{.525mL}} $ .
Thus, $ {\text{Molarity = }}\dfrac{{0.0278}}{{0.0925}} = 0.30 $ .
As a result, the molarity of the solution is $ 0.3{\text{M}} $ .
The correct answer is Option A.

Note:
Intravenous sugar solution, also known as dextrose solution, is a mixture of dextrose (glucose) and water. It is used to treat low blood sugar or water loss without electrolyte loss. Water loss without electrolyte loss may occur in fever, hyperthyroidism, high blood calcium, or diabetes insipidus. It is also used in the treatment of high blood potassium, diabetic ketoacidosis, and as part of parenteral nutrition. It is given by injection into a vein.
Administering a 5% sugar solution peri- and postoperatively usually achieves a good balance between starvation reactions and hyperglycemia caused by sympathetic activation. A 10% solution may be more appropriate when the stress response from the reaction has decreased, after approximately one day after surgery. After more than approximately two days, a more complete regimen of total parenteral nutrition is indicated.