Question

Prepare a 200 grams of a 0.15 Molal Glucose (C6H12O6) solutiona. What weight of Solute is neededb. What weight Water is needed

Answer 1

According to the definition, the molality is the number of the moles of the solute $n$ in mol, divided by the mass of the solvent $m$ in kg:

$b =\frac{n}{m}.$

The mass of the solution is the sum of the masses of the solvent and the solute :

$m_{solution}=m_{solvent}+m_{solute}$ .

The mass of the solute can be expressed as the product of its number of the moles and its molar mass:

$m_{solute}=nM$ .

The molar mass of glucose is 180.156 g/mol. If we link and rearrange these three equations, we end up with the following expression :

$1/b = m_{solution}/n - M$ .

Using this relation , we show that in 200 g of the solution , there is 0.0292 mol of glucose, or 5.26 g of glucose. Finally, 200 g -5.26 g = 194.74 g of water is needed to prepare the solution.

Answer: 5.26 g of glucose and 194.74 g of water are needed in order to prepare 200 g of 0.15 molal glucose solution.