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

Answer:

a) 63,0%

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

c) 13,8 M

Explanation:

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

$\frac{1,704 kg HNO_3}{2,704 kg Solution}$ = 63,0%

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

Pounds HNO₃:

1,704 kg $\frac{2,20462 pounds}{1 kg}$ = 3,7567 pounds

Cubic foot:

2,704 kg $\frac{1 L}{1,382 kg}$x$\frac{1 cubic foot }{28,3168 L}$ = 0,069 ft³

Thus:$\frac{3,7567 pounds}{0,069ft^3} =$ = 54,4 Pounds HNO₃ per cubic foot of solution

c) Moles of HNO₃:

1704 g HNO₃ $\frac{1 mol HNO_3}{63,01 g }$ = 27,04 moles

Liters of solution:

2,704 kg $\frac{1 L}{1,382 kg}$ = 1,96 L of solution

Molarity:

$\frac{27,04 mol}{1,96 L}$ = 13,8 M