Question

A 300 g solution of density 1.5 g/ml is prepared by adding 90 g of glucose ($C_6H_{12}O_6$) in water. Find molarity of this solution.

• A. 5/3
• B. 5/2
• C. 2/5
• D. 3/5

Answer 1

Answer: (B)

The molarity of the solution is 5/2 M.

Answer 2

To find the molarity of the glucose solution:

1. Calculate the molar mass of glucose ($C_6H_{12}O_6$):

   • Molar mass = (6 × 12) + (12 × 1) + (6 × 16) = 72 + 12 + 96 = 180 g/mol

2. Calculate the moles of glucose:

   • Moles of glucose = $\frac{Mass \, of \, glucose}{Molar \, mass \, of \, glucose} = \frac{90 \, g}{180 \, g/mol} = 0.5 \, mol$

3. Calculate the volume of the solution:

   • Volume = $\frac{Mass}{Density} = \frac{300 \, g}{1.5 \, g/ml} = 200 \, ml$

4. Convert the volume from milliliters to liters:

   • Volume = $\frac{200 \, ml}{1000 \, ml/L} = 0.2 \, L$

5. Calculate the molarity:

   • Molarity = $\frac{Moles \, of \, glucose}{Volume \, of \, solution \, in \, liters} = \frac{0.5 \, mol}{0.2 \, L} = 2.5 \, M = \frac{5}{2} \, M$

Therefore, the molarity of the solution is $\frac{5}{2}$ M.