Answer:
Step-by-step explanation:
To calculate the molality of the solution, we first need to determine the number of moles of solute (CH3OH) in 1 kg of the solvent (H2O).
The density of the solution is given as 0.858 g/mL. This means that 1 L (1000 mL) of the solution has a mass of 858 g.
We can assume that the mass of the solvent (water) is 1000 g (i.e., 1 kg). Therefore, the mass of the solute (methanol) in 1 L of the solution is:
mass of CH3OH = mass of solution - mass of H2O
mass of CH3OH = 858 g - 1000 g = -142 g
The negative value indicates that there is no methanol in 1 L of pure water. Therefore, we need to use a smaller volume of the solution that contains a measurable amount of methanol. For example, let's consider 1 L of the solution:
mass of CH3OH in 1 L = 0.858 kg/L × 0.203 mol/L × 32.04 g/mol = 5.58 g
Now, we can calculate the number of moles of CH3OH in 1 kg of H2O:
moles of CH3OH = 5.58 g / 32.04 g/mol = 0.1745 mol
The molality of the solution is then:
molality = moles of solute / mass of solvent (in kg)
mass of 1 L of water = 1000 g / 1000 = 1 kg
molality = 0.1745 mol / 1 kg = 0.1745 m
Therefore, the molality of the 20.3 m CH3OH solution is 0.1745 m.