Question

Given a water solution that contains 1.704 kg of HNO:/kg H:O and has a specific gravity of 1.382 at 20 °C, express the composition in the following ways: (a) Weight percent HNO (b) Pounds HNO3 per cubic foot of solution at 20 °c (c) Molarity (gmoles of HNOs per liter of solution at 20 °C)

Answer 1

a) 63,0%

b) 54,4 Pounds HNO₃ per cubic foot of solution

c) 13,8 M

Step-by-step explanation:

a) Weight percent is the ratio solute:solution times 100:

`(1,704 kg HNO_3)/(2,704 kg Solution)` = 63,0%

b) Pounds HNO₃ per cubic foot of solution at 20 °c

Pounds HNO₃:

1,704 kg
`(2,20462 pounds)/(1 kg)` = 3,7567 pounds

Cubic foot:

2,704 kg
`(1 L)/(1,382 kg)`x
`(1 cubic foot )/(28,3168 L)` = 0,069 ft³

Thus:
`(3,7567 pounds)/(0,069ft^3) =` = 54,4 Pounds HNO₃ per cubic foot of solution

c) Moles of HNO₃:

1704 g HNO₃
`(1 mol HNO_3)/(63,01 g )` = 27,04 moles

Liters of solution:

2,704 kg
`(1 L)/(1,382 kg)` = 1,96 L of solution

Molarity:

`(27,04 mol)/(1,96 L)` = 13,8 M