A 2.050×10−2 MM solution of glycerol (C3H8O3C3H8O3) in water is at 20.0∘C∘C. The sample was created by dissolving a sample of C3H8O3C3H8O3 in water and then bringing the volume up to 1.000 LL. It was determined that the volume of water needed to do this was 999.1 mLmL . The density of water at 20.0∘C∘C is 0.9982 g/mLg/mL. Part A Calculate the molality of the glycerol solution. Express your answer to four significant figures and include the appropriate units.

[Glycerol] = 0.0205 mol/kg

Explanation:

Molality means concentration. This sort of concentration indicated the moles of solute which are contained in 1kg of solvent.

We start from molarity which are the moles of solute, that are contained in 1L of solution.

2.05×10⁻² mol/L are contained in 1 L of solution.

We must assume, that volume of solvent is 999.1 mL so let's determine the mass of solvent, by density

999.1 mL . 0.9982 g/7mL = 997.3 g

For molality we need mass in kg, so let's convert the mass of solvent

997.3 g . 1kg /1000g = 0.9973 kg.

Molality → mol/kg → 2.05×10⁻² mol / 0.9973 kg = 0.0205 m

the density of gold = 19.3 g/cm3

the relation between density and mass and volume is

density = mass / volume

given :

mass of gold = 13 g

density = 13g / volume = 19.3

volume = 13 /19.3 = 0.67 cm^3

thus volume of gold nugget = 0.67 cm^3