Glucose makes up about \(0.10 \%\) by mass of human blood. Calculate this concentration in (a) ppm, (b) molality. (c) What further information would you need to determine the molarity of the solution?

To convert the mass percent concentration to parts per million (ppm), we need to multiply the mass percent by 10,000 (since there are 1,000,000 parts in a million and 100 parts in a percent): ppm = mass percent × 10,000 ppm = 0.10% × 10,000 ppm = 1000 Therefore, the concentration of glucose in human blood is approximately 1000 ppm.

To calculate molality, we need to find the moles of glucose and mass of solvent (water) in 1L solution. Since the given problem states that glucose makes up 0.10% by mass of human blood, we will consider 1L of blood with a density of \(1 g/mL\), meaning total mass of blood is \(1000g\). First, let's find the mass of glucose in 1000g blood, mass of glucose = mass percent × total mass mass of glucose = (0.10 / 100) × 1000 = 1g Next, we will find the mass of the solvent (water): mass of water = total mass - mass of glucose mass of water = 1000 - 1 = 999g Now let's find the moles of glucose. Given the molecular formula of glucose, C6H12O6, its molar mass is \(6 × 12.01(g/mol) + 12 × 1.01(g/mol) + 6 × 16.00(g/mol) = 180.16(g/mol)\). moles of glucose = mass / molar mass moles of glucose = 1 / 180.16 = 0.00555 mol Finally, let's find the molality, molality = moles of solute / mass of solvent (in kg) molality = 0.00555mol / 0.999kg = 0.00556 \(mol/kg\) Therefore, the molality of glucose in human blood is approximately 0.00556 \(mol/kg\).

To determine the molarity of the glucose solution, we would need to know the volume of blood (solvent) in which the given moles of solute (glucose) are dissolved. Since molarity is the moles of solute per liter of solution, knowing the volume of the solution would allow us to calculate the molarity.