Question

If 18 g of glucose is present in 1000 g of solvent, the solution is said to be:
A.1 molar
B.0.1 molar
C.0.5 molal
D.0.1 molal

Answer 1

To answer this question, we need to first understand the difference between molarity and molality. Molarity is the number of moles of solute in one litre of the solution. Molality is the number of moles of solute per kilogram of the solvent. Thus we can say that by looking at the given values we can find the correct option.

Complete answer:
We are given 18 grams of glucose. We know that one mole of glucose has a molar mass of 180 grams per mole. Thus we can say that the total number of moles of glucose in the given sample is
$\Rightarrow No.{\text{ }}of{\text{ }}moles = \dfrac{{Given{\text{ }}mass}}{{Molar{\text{ }}mass}}$
Substituting the known values in the given equation we get:
$\Rightarrow No.{\text{ }}of{\text{ }}moles = \dfrac{{18g}}{{180g/mol}}$
Thus by simplification we get:
$\Rightarrow No.{\text{ }}of{\text{ }}moles = 0.1{\text{ }}moles$
We are also provided with 1000 grams of the solvent. We can say that 1000 grams of the solvent means 1 kilogram of the solvent.
Thus we can say the given data correspond to molality, as we are given with weight of the solvent.
$\Rightarrow Molality = \dfrac{{No.{\text{ }}of{\text{ }}moles{\text{ }}of{\text{ }}solute}}{{Mass{\text{ }}of{\text{ }}solvent{\text{ }}in{\text{ }}kg}}$
We can substitute all the values in this equation and thus obtain the equation below:
$\Rightarrow Molality = \dfrac{{0.1{\text{ }}moles}}{{1{\text{ }}kg}}$
$\Rightarrow Molality = 0.1{\text{ }}molal$
Thus we can say that the correct option is (D).

Note:
In this question, we could easily eliminate option (A) and (B) because the values given in the question correspond only to molarity. Do not confuse molarity and molality when it is given in a question. There are relations that can help in the interconversion of both these quantities. The relation is as given below:
$\Rightarrow \dfrac{1}{m} = \dfrac{d}{M} - \dfrac{{{m_B}}}{{1000}}$
m is the molality
M is the molarity
D is the density of the solution
${m_B}$ is the molar mass of the solute