Answer:
The molality of the solution is m = m(CH3OH) + m(NaBr) = 31.21 mol/kg + 0.12 mol/kg = 31.33 mol/kg (rounded to two decimal places).
Step-by-step explanation:
To find the molality of a solution, we need to first calculate the moles of solute per kilogram of solvent.
Step 1
Calculate the moles of each substance.
The molar mass of methanol (CH3OH) is
1 x 12.01 g/mol (C) + 4 x 1.01 g/mol (H) + 1 x 16.00 g/mol (O) = 32.04 g/mol
The number of moles of methanol is
1330 g / 32.04 g/mol = 41.52 mol
The molar mass of sodium bromide (NaBr) is
1 x 22.99 g/mol (Na) + 1 x 79.90 g/mol (Br) = 102.89 g/mol
The number of moles of sodium bromide is
16.6 g / 102.89 g/mol = 0.1612 mol
Step 2
Calculate the mass of the solvent in kilograms.
The total mass of the solution is
1330 g + 16.6 g = 1346.6 g
To find the mass of the solvent, we subtract the mass of the solute
1346.6 g - 16.6 g = 1330 g
The mass of the solvent is 1.33 kg.
Step 3
Calculate the molality.
The molality (m) is defined as the number of moles of solute per kilogram of solvent
m = moles of solute / mass of solvent in kg
m(CH3OH) = 41.52 mol / 1.33 kg = 31.21 mol/kg
m(NaBr) = 0.1612 mol / 1.33 kg = 0.12 mol/kg
Therefore, the molality of the solution is
m = m(CH3OH) + m(NaBr) = 31.21 mol/kg + 0.12 mol/kg = 31.33 mol/kg (rounded to two decimal places).