Question

$ 36g $ of glucose (molar mass $ =\text{ }180\text{ }g/mol $) is present in $ 500g $ of water, the molality of the solution is
(A) $ 0.2 $
(B) $ 0.4 $
(C) $ 0.8 $
(D) $ 1.0 $

Answer 1

Hint: We know that for solving the given question you should have knowledge about the following terms Molality, Molarity and Mole fraction. - Molarity (M) is defined as a number of moles of solute dissolved in one litre of solution. Molality (m) is defined as the number of moles of the solute per kilogram (kg) of the solvent. Mole fraction represents the number of moles of a particular component in a mixture divided by the total number of moles in the given mixture. It can be represented by X.

Complete answer:
We also know that $ 10% $ by w/w solution of glucose means there is $ 10g $ of glucose and $ 90g $ of water in $ 100g $ of solution. And the Molality is calculated by no. of moles of solute/ mass of solvent in Kilograms. The concentrations which are expressed in molality are used while studying properties of solutions Formula Used: Molality no. of moles of solute/ mass of solvent
Some uses of molality are: The concentrations which are expressed in molality are used while studying properties of solutions linked to vapor pressure and temperature changes. Molality is also used because its value does not change with variations in temperature. The volume of a solution, on the other hand, is somewhat dependent upon temperature.
 $ Molality(m)=\dfrac{Moles\text{ }of\text{ }solute\times 1000}{Moles\text{ }of\text{ }solvent(g)}=\dfrac{36\times 1000}{180\times 500}=0.4m $
Therefore, correct answer is option B i.e. $ 36g $ of glucose (molar mass $ =\text{ }180\text{ }g/mol $) is present in $ 500g $ of water, the molality of the solution is $ 0.4m. $

Note:
Remember that Molarity is expressed by “M” whereas molality is expressed by “m” and these both express the concentration of a chemical solution. And always remember that $ 10% $ by w/w solution of glucose means there is of glucose and of water $ 100g $ in solution. It will help in solving the problem given.