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): \(2.06\)
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)

Cobalt chloride when dissolved in water forms pink colored complex $X$ which has octahedral geometry. This solution on treating with cone $HCl$ forms deep blue complex, $\underline{Y}$ which has a $\underline{Z}$ geometry $X, Y$ and $Z$, respectively, are
| Sample | Van't Haff Factor |
|---|---|
| Sample - 1 (0.1 M) | \(i_1\) |
| Sample - 2 (0.01 M) | \(i_2\) |
| Sample - 3 (0.001 M) | \(i_2\) |
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)
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: