Question

A 2.550

Answer 1

1.

Molality = moles of solute / Mass of solvent in Kg

Given, 2.550 x 10^-2 M solution. = 2.550 x 10^-2 moles / L of solution.

So, moles = 2.550 x 10^-2

Given, volue of water needed = 999.3 mL

Mass of solvent = density x volume = 0.9982 x 999.3 g = 997.501 g = 0.99750 Kg

Molality = 2.550 x 10^-2 / 0.99750 = 0.02556 m

Molality = 0.02556 m

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2.

Mole fraction of salt = Moles of salt / (Moles of salt + Moles of solvent)

Moles of salt = 2.550 x 10^-2 mol

Moles of solvent = Mass / Molar mass = 997.50 g / 18 g/mol = 55.41 mol

Mole fraction of salt = 2.550 x 10^-2 / (2.550 x 10^-2 + 55.41)

=4.591 x 10^-4

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3.

Mass percent = (mass of solute / Mass of solution) x 100

Moles of solute = 2.550 x 10^-2 mol

Mass of solute = Moles x molar mass = 2.550 x 10^-2 mol x 58.44 g/mol = 1.4902 g

Mass of solution = 997.501 g

Mass percent = (1.4902 / 997.501) x 100

= 0.1494 %

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4.

ppm = mass of solute (mg) / mass of solution in Kg

Mass of solute = 1.4902 g = 1490.2 mg

Mass of solution = 997.50 g = 0.99750 Kg

Concentration = 1490.2 mg / 0.99750 Kg = 1494 ppm