Out of \(18 g\) of glucose only \(4.5 g\) is dissolved in water to obtain \(40 mL\) of its aqueous solution. Find the molarity of this glucose solution

0.625 M

The correct answer is option 4. 0.625 M.

To find the molarity (\(M\)) of the glucose solution, we first need to calculate the number of moles of glucose (\(n\)) that are dissolved in the solution, and then divide by the volume of the solution in liters (\(V\)).

Given:

Mass of glucose (\(m\)) = \(4.5 \, g\)

Molar mass of glucose = \(180.16 \, g/mol\) (approximately, for simplicity, the molecular formula of glucose is \(C_6H_{12}O_6\))

Volume of solution (\(V\)) = \(40 \, mL = 0.040 \, L\)

First, let's calculate the number of moles of glucose (\(n\)) using the formula:

\( n = \frac{m}{M} \)

Where:

\( m \) is the mass of glucose (in grams)

\( M \) is the molar mass of glucose (in grams per mole)

Substituting the given values:

\(n = \frac{4.5 \, g}{180.16 \, g/mol} \)

\( n \approx 0.02497 \, mol \)

Now, let's calculate the molarity (\(M\)) using the formula:

\( M = \frac{n}{V} \)

Where:

\( n \) is the number of moles of solute (in moles)

\( V \) is the volume of the solution (in liters)

Substituting the given values:

\( M = \frac{0.02497 \, mol}{0.040 \, L} \)

\( M \approx 0.62425 \, M \)

Rounding off to three significant figures, the molarity of the glucose solution is approximately \(0.625 \, M\).

So, the correct answer is option 4. \(0.625 \, M\).