What mass of glucose (C6H12O6) must dissolve in 400.0 g of ethanol to make a 1.60 m solution?

To determine the mass of glucose (C6H12O6) required to make a 1.60 m solution in 400.0 g of ethanol, we need to calculate the number of moles of glucose required.

Step 1: Calculate the number of moles of ethanol.
Since we are given the mass of ethanol (400.0 g) and the molar mass of ethanol is 46.07 g/mol, we can calculate the number of moles using the formula:

moles of ethanol = mass of ethanol / molar mass of ethanol

moles of ethanol = 400.0 g / 46.07 g/mol

Step 2: Calculate the volume of the solution.
To calculate the volume of the solution, we need to use the formula:

volume of solution = mass of ethanol / density of ethanol

Assuming the density of ethanol is 0.789 g/mL, we can substitute the values:

volume of solution = 400.0 g / 0.789 g/mL

Step 3: Calculate the moles of glucose.
Since the concentration (molarity) is given as 1.60 m (moles per liter), we can calculate the total moles of glucose using the formula:

moles of glucose = molarity × volume of solution

moles of glucose = 1.60 mol/L × (volume of solution in liters)

Step 4: Convert moles of glucose to grams.
Since the molar mass of glucose is 180.16 g/mol, we can calculate the mass of glucose using the formula:

mass of glucose = moles of glucose × molar mass of glucose

Finally, we substitute the calculated values into the equation to find the mass of glucose:

mass of glucose = (1.60 mol/L × (volume of solution in liters)) × (180.16 g/mol)

To determine the mass of glucose (C6H12O6) required to make a 1.60 m solution in ethanol, we need to consider the definition of molarity and use some conversion factors.

Molarity (M) is defined as the number of moles of solute per liter of solution. In this case, the molarity is given as 1.60 m, which means there are 1.60 moles of glucose per liter of solution.

First, we need to convert the given mass of ethanol (400.0 g) to liters. To do this, we need to know the density of ethanol. The density of ethanol at room temperature is approximately 0.789 g/mL.

Density = Mass / Volume

Rearranging the equation, we can calculate the volume (V) in liters:

Volume (L) = Mass (g) / Density (g/mL)

Volume (L) = 400.0 g / 0.789 g/mL = 506.34 mL

Next, we convert the volume from milliliters to liters:

Volume (L) = 506.34 mL × 1 L / 1000 mL = 0.50634 L

Now we can calculate the number of moles of glucose required:

Moles = Molarity (M) × Volume (L)

Moles = 1.60 mol/L × 0.50634 L = 0.8117 mol

Finally, we need to convert the moles of glucose to grams using the molar mass of glucose (C6H12O6):

Molar mass of glucose (C6H12O6) = 6(12.01 g/mol) + 12(1.01 g/mol) + 6(16.00 g/mol)
= 180.18 g/mol

Mass (g) = Moles × Molar mass (g/mol)

Mass (g) = 0.8117 mol × 180.18 g/mol = 146.84 g

Therefore, approximately 146.84 grams of glucose (C6H12O6) must dissolve in 400.0 g of ethanol to make a 1.60 m solution.

3.225

m = molality = mols/kg solvent.

1.60 = mols glucose/0.4000). Solve for mols glucose, then mols = grams/molar mass. You know molar mass and mols, solve for grams.