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melamori03 [73]
3 years ago
15

The concentrated sulfuric acid we use in the laboratory is 98.0% sulfuric acid by weight. Calculate the molality and molarity of

concentrated sulfuric acid if the density of the solution is 1.83 g cm-3.
Chemistry
1 answer:
timama [110]3 years ago
7 0

Answer : The molarity and molality of the solution is, 18.29 mole/L and 499.59 mole/Kg respectively.

Solution : Given,

Density of solution = 1.83g/cm^3=1.83g/ml

Molar mass of sulfuric acid (solute) = 98.079 g/mole

98.0 % sulfuric acid by mass means that 98.0 gram of sulfuric acid is present in 100 g of solution.

Mass of sulfuric acid (solute) = 98.0 g

Mass of solution = 100 g

Mass of solvent = Mass of solution - Mass of solute = 100 - 98.0 = 2 g

First we have to calculate the volume of solution.

\text{Volume of solution}=\frac{\text{Mass of solution}}{\text{Density of solution}}=\frac{100g}{1.83g/ml}=54.64ml

Now we have to calculate the molarity of solution.

Molarity=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{volume of solution}}=\frac{98.0g\times 1000}{98.079g/mole\times 54.64ml}=18.29mole/L

Now we have to calculate the molality of the solution.

Molality=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Mass of solvent}}=\frac{98.0g\times 1000}{98.079g/mole\times 2g}=499.59mole/Kg

Therefore, the molarity and molality of the solution is, 18.29 mole/L and 499.59 mole/Kg respectively.

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