How many kg of wet NaOH containing 12% water are required to prepare 60 litre of 0.5N solution?
Let's assume for the moment the water is not there.
How many mols do we want. That is M x L = mols = 0.5 x 60 = ? (I've used molarity since mols and equivalents for NaOH are the same).
How many grams is that?
grams = mols x molar mass IF IT WERE PURE STUFF BUT IT'S ONLY 100-12 = 88% NaOH. Therefore, we need grams from above/0.88 = grams of the wet stuff.
To solve this question, we need to follow a few steps.
Step 1: Calculate the number of moles of NaOH required.
The number of moles can be calculated using the formula:
Moles = Volume x Molarity
Here, the volume is given as 60 liters and the molarity is given as 0.5N. The unit 'N' represents the normality, which is defined as the number of equivalents of a solute per liter of solution. For NaOH, 1N means 1 equivalent per liter.
Since we want to calculate in moles, we need to convert 0.5N to the molar concentration. To do this, we need to know the molecular weight of NaOH. The molecular weight of NaOH is 23 (for Na) + 16 (for O) + 1 (for H) = 40 g/mol.
To convert from N to Molarity (M), we divide by the number of equivalents. For NaOH, 1N equals 1 equivalent, so 0.5N is equal to 0.5 moles/liter or 0.5 M.
Therefore, the number of moles of NaOH required is:
Moles = 60 L x 0.5 M = 30 moles
Step 2: Calculate the weight of NaOH required.
To calculate the weight of NaOH required, we need to consider that the given NaOH is wet and contains 12% water. This means that 88% of the weight is actual NaOH, while the remaining 12% is water.
Let's assume the weight of the wet NaOH is W kg. Then, the weight of the actual NaOH present in W kg of wet NaOH is 88% of W kg, which can be written as 0.88W kg.
To find the weight of the actual NaOH required to get the desired number of moles, we need to use the molecular weight of NaOH again. The molecular weight is 40 g/mol, which is equal to 0.04 kg/mol (because there are 1000 grams in a kilogram).
Therefore, the weight of NaOH required is:
(0.88W kg) = (30 moles) x (0.04 kg/mol)
0.88W = 1.2
W = 1.2 / 0.88
W ≈ 1.36 kg
So, approximately 1.36 kg of wet NaOH containing 12% water is required to prepare a 60-liter solution with a molarity of 0.5N.