Calculate the molality of of a sol'n whose density is 1.15 g/ml

Is this a problem you made up or a homework question from class or your text?

oh I'm sorry it's a typo

The question is this:

Calculate the molality of a 0.410 molar aqueous glucose solution whose density is 1.15 g/mol

To calculate the molality of a solution, we first need to know the mass of solute dissolved in a given mass of solvent.

Molality (m) is defined as the number of moles of solute (n) dissolved in 1 kilogram (1000 grams) of solvent. It is expressed in moles of solute per kilogram of solvent.

In this case, we do not have information about the nature of the solute or the solute's mass. Therefore, we cannot directly calculate the molality of the solution based solely on the density.

We would need additional information such as the identity and mass of the solute to calculate the molality.

0.410 M glucose means 0.410 mols glucose/L soln. The molar mass of glucose is approximately 180 g/mol which means 0.410 is 0.410 x 180 = about 74g. (You need to go through all of these estimates and use better values.)

The mass of 1000 mL of the soln is
1.15 g/mL x 1000 = 1150 g which means the mass of the solvent is 1150-74 = ?. Then m = mols/kg solvent.