Answer :
Answer:
25.4 M
Explanation:
From the question, 19.4 M implies that there are 19.4 moles of NaOH in 1000ml solution.
Hence
Mass of the solution = density × volume
= 1000×1.54 = 1540 g
Mass of pure NaOH = 19.4 moles × 40g/mol = 776 g
Mass of water = 1540 - 776 = 764 g
Therefore;
Molarity of solution= 19.4/764 × 1000 = 25.4 M
The molality of concentrated sodium hydroxide, which is 19.4 M and has a density of 1.54 g/mL, is 25.4 mol/kg.
What is molality?
Molality (b) is a unit of concentration, equal to the number of moles of solute per kilogram of solvent.
To solve this problem, let's suppose we have 1 L of solution, although we would get to the same result with any volume.
- Step 1: Calculate the moles and mass of solute in 1 L of solution.
The molarity of the solution is 19.4 M, that is, there are 19.4 moles of solute in 1 L of solution.
We can convert 19.4 moles to mass using the molar mass of NaOH.
19.4 mol × 40.0 g/the = 776 g
- Step 2: Calculate the mass corresponding to 1 L of solution.
1 L (1000 mL) of the solution has a density of 1.54 g/mL.
1000 mL × 1.54 g/mL = 1540 g
- Step 3: Calculate the mass of solvent (water) is 1 L of solution
In 1540 g of solution, there are 776 g of solute. The rest is water.
1540 g - 776 g = 764 g = 0.764 kg
- Step 4: Calculate the molality of the solution.
b = 19.4 mol / 0.764 kg = 25.4 mol/kg
The molality of concentrated sodium hydroxide, which is 19.4 M and has a density of 1.54 g/mL, is 25.4 mol/kg.
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