Question

A gas has a Henry's law constant of 6.10 × 10 − 4 M/atm at 21.1 °C. What volume of solution is needed to completely dissolve 4.56 L of the gas at 1335 torr and 21.1 °C?

Answer 1

Answer:

311 L

Explanation:

Henry's law states that the solubility of a gas is directly proportional to the pressure of that gas done above the liquid that it's dissolved:

S = H*P

Where S is the concentration (mol/L), H is Henry's law constant, and P is the pressure.

P = 1335 torr * 1atm/760torr = 1.76 atm

S = 6.10x10⁻⁴ * 1.76

S = 1.07x10⁻³ M

By the ideal gas law:

PV = nRT

Where P is the pressure, V is the volume of the gas, n is the number of moles, R is the ideal gas constant (0.082 atm.L/mol.K), and T is the temperature (21.1°C = 294.25 K)

1.76*4.56 = n*0.082*294.25

24.1285n = 8.0256

n = 0.33262 mol

The concentration (S) is the number of moles (n) divided by the volume of the solution (Vl) so:

S = n/Vl

Vl = n/S

Vl = 0.33262/1.07x10⁻³

Vl = 311 L

Answer 2

The volume of the solution needed to completely dissolve the gas is 310.28 L.

The solubility of the gas can be determined by applying Henry's law as follows;

S = HP

where;

H is Henry's law constant = 6.1 x 10⁻⁴ M/atm

P is the pressure of the gas, = 1335 torr = 1.757 atm

$S = 6.1 \times 10^{-4} \times 1.757 \\\\S = 1.07 \times 10^{-3} \ M$

Apply ideal gas law to determined the number of moles of the gas at the given temperature;

$PV = nRT\\\\n = \frac{PV}{RT} \\\\n = \frac{1.757 \ atm \ \times \ 4.56\ L}{(0.0821 \ L.atm/mol.K) \ \times\ ( 21.1 + 273)K} \\\\n = 0.332 \ mole$

The volume of the solution needed to completely dissolve the gas at the given conditions is calculated as follows;

$n = SV\\\\V = \frac{n}{S} \\\\V = \frac{0.332}{1.07 \times 10^{-3} } \\\\V = 310.28 \ L$

Learn more about Henry's law here: https://brainly.com/question/14665214