How many grams of KOH will be needed to boil a liter of cold water?
To determine the amount of KOH needed to boil a liter of cold water, we need to consider the boiling point elevation caused by adding the solute, which in this case is KOH (potassium hydroxide). The boiling point elevation is a colligative property that depends on the concentration of the solute.
To calculate the amount of KOH required, we'll use the formula:
ΔT = K * m,
where:
- ΔT is the boiling point elevation,
- K is the molal boiling point elevation constant (which is specific to the solvent, in this case, water),
- m is the molality of the solute (the number of moles of KOH per kilogram of water).
The boiling point elevation constant for water is approximately 0.512 °C/m.
First, we need to determine the molality of the KOH solution. We know that 1 liter of water weighs approximately 1000 grams. So, for every kilogram of water, we need to dissolve the molar mass of KOH (which is roughly 56.1 grams) in order to obtain a molar concentration.
molality (m) = moles of solute / mass of solvent (kg)
Since we're using 1 liter of water, the mass of the solvent is 1 kg (or 1000g in this case).
m = (56.1 g) / (1000 g) = 0.0561 mol/kg
Now, we can calculate the boiling point elevation:
ΔT = K * m
ΔT = (0.512 °C/m) * (0.0561 mol/kg) ≈ 0.0288 °C
The boiling point elevation caused by adding KOH to 1 liter of cold water is approximately 0.0288 °C.
Please note that this calculation assumes ideal behavior, and there may be practical factors (such as non-ideal solutions or heat losses) that could cause deviations from the theoretical value.