Question

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

Answer 1

Answer:

b

Explanation:

0.15=(200×1000)/(180.156×7400)

0.15= mols of glucose soln

200= weight of glucose

180.156=molecular weight of glucose

7400= lt of water

Answer 2

According to the molality, 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 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.