To find the molality of the solution, we need to understand the details provided and make some calculations:
Thus, the molality of the solution is nearest to 2.06.
Weight of D-glucose in water =\( 250\) \(g\)
∴ Weight of carbon in D-glucose
= \(\frac{250}{180}×72 = 100\) \(g\)
Percentage of carbon in the aqueous solution of glucose is
= \(10.8\%\)
∴ Weight of the solution is = \(925.93\)
∴ Molality of D-glucose is
=\(\frac{\frac{ 250}{180}}{(925.93 - 250)}×1000\)
= \(\frac{250}{180×675.93} ×1000\)
= \(2.06\)
Hence, the correct option is (B): \(2.06\)
A substance 'X' (1.5 g) dissolved in 150 g of a solvent 'Y' (molar mass = 300 g mol$^{-1}$) led to an elevation of the boiling point by 0.5 K. The relative lowering in the vapour pressure of the solvent 'Y' is $____________ \(\times 10^{-2}\). (nearest integer)
[Given : $K_{b}$ of the solvent = 5.0 K kg mol$^{-1}$]
Assume the solution to be dilute and no association or dissociation of X takes place in solution.
Which one of the following graphs accurately represents the plot of partial pressure of CS₂ vs its mole fraction in a mixture of acetone and CS₂ at constant temperature?

In the given figure, the blocks $A$, $B$ and $C$ weigh $4\,\text{kg}$, $6\,\text{kg}$ and $8\,\text{kg}$ respectively. The coefficient of sliding friction between any two surfaces is $0.5$. The force $\vec{F}$ required to slide the block $C$ with constant speed is ___ N.
(Given: $g = 10\,\text{m s}^{-2}$) 
A solution is a homogeneous mixture of two or more components in which the particle size is smaller than 1 nm.
For example, salt and sugar is a good illustration of a solution. A solution can be categorized into several components.
The solutions can be classified into three types:
On the basis of the amount of solute dissolved in a solvent, solutions are divided into the following types: