Question

D5W refers to one of the solutions used as intravenous fluid. It is a 5% by mass solution of dextrose, $C_6{H}_{12}{O}_6$ in water. The density of D5W is 1.08g/ml. The molarity of the solution is
A) 0.3M
B) 0.6M
C) 0.28M
D) 0.26M

Answer 1

Hint: The answer to this question is based on the fact that D5W solution is nothing but 5% solution of dextrose where the molarity of this solution is based on the formula that is given by $Molarity={\displaystyle \frac{m}{V}}$

Complete step – by – step solution:
We have studied in our previous chapters of chemistry that deals with the concepts of physical chemistry which includes calculation of molality, molarity and mole fraction and several other terms related to it.
Let us now work on calculation of molarity of the dextrose solution.
- D5W solution or 5% solution of dextrose is a form of glucose in which the name is expanded as ‘dextrose 5% in water’ and this solution is used as intravenous liquid.
- This solution is injected into the vein through an IV which replaces the lost fluids and this helps to provide the carbohydrates in the body.
Given data is as follows,
% by mass = 5
This means that 5 g of $C_6{H}_{12}{O}_6$in 100 grams of solution.
Moles of $C_6{H}_{12}{O}_6={\displaystyle \frac{5}{180}}=0.027778$
We know that volume of the solution is given by,
$V={\displaystyle \frac{mass}{density}}$
$$\Rightarrow V={\displaystyle \frac{100}{1.08}}=92.5925ml$$
Now, molarity of the solution is given by,
$Molarity={\displaystyle \frac{m}{V}}$
where m is the number of moles of $C_6{H}_{12}{O}_6$
V is the volume of the solution
By substituting the values, we have
$$Molarity={\displaystyle \frac{0.027778}{0.0925g}}$$ [since it should be in form of 1 litre of solution]
Thus, molarity = 3

Therefore the correct answer is option A) 0.3M.

Note: Note that molarity and molality are different and do not be confused with the formula as molarity is the measure of number of moles of solute present in 1 $d{m}^3$ of solution whereas molality is the measure of number of moles of solute present in 1 kg of the solvent.