Calculate the molality of 0.1 molal NaOH solution having density 1 gra...
Molality of NaOH Solution
To calculate the molality of a solution, we need to know the amount of solute (in moles) and the mass of the solvent (in kilograms). In this case, we have a 0.1 molal NaOH solution with a density of 1 gram per ml. Let's break down the calculation step by step:
Step 1: Convert density to mass of solvent
The density of the solution is given as 1 gram per ml. Since the density is in grams per ml, we can directly use it as the mass of the solution. However, we need to determine the mass of the solvent (water) by subtracting the mass of the solute (NaOH) from the mass of the solution.
Step 2: Determine the mass of solute (NaOH)
The molality of a solution is defined as the number of moles of solute per kilogram of solvent. Since the molality is given as 0.1 molal, we know that there are 0.1 moles of NaOH in 1 kilogram of water.
The molar mass of NaOH is calculated as follows:
Na (sodium) = 22.99 g/mol
O (oxygen) = 16.00 g/mol
H (hydrogen) = 1.01 g/mol
Molar mass of NaOH = 22.99 + 16.00 + 1.01 = 40.00 g/mol
Therefore, the mass of 0.1 moles of NaOH is:
0.1 mol × 40.00 g/mol = 4.00 grams
Step 3: Determine the mass of solvent (water)
To find the mass of water, we subtract the mass of NaOH from the mass of the solution. Since the density of the solution is given as 1 gram per ml, the mass of the solution is equal to the volume of the solution.
Let's assume we have 1000 ml of the solution. Therefore, the mass of the solution is 1000 grams. Subtracting the mass of NaOH (4.00 grams) from the mass of the solution, we get the mass of the solvent (water) as:
1000 g - 4.00 g = 996 grams
Step 4: Calculate the molality
Finally, we can calculate the molality of the NaOH solution using the equation:
Molality (m) = moles of solute / mass of solvent (in kg)
We know that the moles of solute (NaOH) is 0.1 moles and the mass of the solvent (water) is 996 grams (0.996 kg).
Plugging these values into the equation:
Molality (m) = 0.1 moles / 0.996 kg = 0.100 mol/kg
Therefore, the molality of the 0.1 molal NaOH solution is 0.100 mol/kg.
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