A solution is prepared by dissolving 17.75 g sulfuric acid, h2so4, in enough water to make 100.0 ml of solution. if the density of the solution is 1.1094 g/ml, what is the molality?

Respuesta :

Answer: The molality of solution is 1.94 m.

Explanation:

We are given:

Mass of sulfuric acid = 17.75 grams

Volume of solution = 100 mL

Density of solution = 1.1094 g/mL

To calculate the mass of solution, we use the equation:

[tex]Density=\frac{Mass}{Volume}[/tex]

Putting values in above equation, we get:

[tex]1.1094g/mL=\frac{\text{Mass of solution}}{100mL}[/tex]

[tex]\text{Mass of solution}=110.94g[/tex]

Mass of solvent = Mass of solution - Mass of solute

Mass of solvent = 110.94 - 17.75 = 93.19 g

To calculate the molality of solution, we use the equation:

[tex]Molality=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}[/tex]

Where,

[tex]m_{solute}[/tex] = Given mass of solute [tex](H_2SO_4)[/tex] = 17.75 g

[tex]M_{solute}[/tex] = Molar mass of solute [tex](H_2SO_4)[/tex] = 98 g/mol

[tex]W_{solvent}[/tex] = Mass of solvent = 93.19 g

Putting values in above equation, we get:

[tex]\text{Molality of }H_2SO_4=\frac{17.75\times 1000}{98\times 93.19}[/tex]

[tex]\text{Molality of }H_2SO_4=1.94m[/tex]

Hence, the molality of solution is 1.94 m.