Answer to Question #260955 in Chemistry for siddik

Question

What mass of potassium nitrate, KNO3, is contained in 400 cm3

of an aqueous

solution of KNO3 that has a concentration of 50.0 ppm K

+

. (Assume the only source of K+

in the

solution is KNO3). Relative Atomic Mass Data: K -39.1, N-14.0, O-16.0

Answer 1

Solution:

Convert ppm to mg/L:

1 ppm = 1 mg L^-1

Hence,

(50.0 ppm) × (1 mg L^-1 / 1 ppm) = 50.0 mg L^-1 K⁺

The molar mass of K⁺ is 39.1 g mol^-1

Volume of an aqueous solution is 400 cm³ or 0.40 L

Hence,

(50.0 mg K⁺ / 1 L) × (1 g / 1000 mg) × (1 mol K⁺ / 39.1 g K⁺) × (0.40 L) = 0.0005115 mol K⁺

KNO₃ → K⁺ + NO₃^-

According to stoichiometry:

Moles of KNO₃ = Moles of K⁺ = 0.0005115 mol

Calculate the mass of KNO₃:

The molar mass of KNO₃ is 101.1g mol^-1

Hence,

(0.0005115 mol KNO₃) × (101.1 g KNO₃ / 1 mol KNO₃) = 0.0517 g KNO₃ = 51.7 mg KNO₃

Answer: The mass of potassium nitrate (KNO₃) is 51.7 mg