Question

. The Henry's law constant for helium gas in water at 30 ∘C is 3.7×10−4M/atm; the constant for N2 at 30 ∘C is 6.0×10−4M/atm. a. If helium gas is present at 2.1 atm pressure, calculate the solubility of this gas. b. If N2 is present at 2.1 atm pressure, calculate the solubility of this gas.2. A solution is made containing 14.7 g of CH3OH in 186 g H2O. a. Calculate the mass percent of CH3OH. b. Calculate the molality of CH3OH.

Answer 1

Explanation:

1) Henry's law states that the amount of gas dissolved or molar solubility of gas is directly proportional to the partial pressure of the liquid.

To calculate the molar solubility, we use the equation given by Henry's law, which is:

$M_{g}=K_H\times p_{g}$

where,

$K_H$ = Henry's constant

$p_{He}$ = partial pressure of gas

a)  $K_H=3.7\times 10^{-4} M/atm$

$p_{He}=2.1 atm$

Putting values in above equation, we get:

$M_{He}=3.7\times 10^{-4} M/atm\times 2.1 atm = 7.77\times 10^{-4} M$

The solubility of helium gas is $7.77\times 10^{-4} M$

b)  $K_H=6.0\times 10^{-4} M/atm$

$p_{N_2}=2.1 atm$

Putting values in above equation, we get:

$M_{He}=6.0\times 10^{-4} M/atm\times 2.1 atm = 1.26\times 10^{-3} M$

The solubility of nitrogen gas is $1.26\times 10^{-3} M$

2)

a) Mass of solute or methanol , m= 14.7 g

Mass of solvent or water , m'= 186 g

Mass of the solution = M = m + m' = 14.7 g + 186 g = 200.7 g

$w/w\%=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 100$

$=\frac{14.7 g}{200.7 g}\times 100=7.32\%$

The mass percent of methanol is 7.32%.

b) Molality is defined as moles of solute per kilograms of solvent.

$m=\frac{\text{Moles of solute}}{\text{Mass of solvent in kg}}$

Moles of methanol = $\frac{14.7 g}{32 g/mol}=0.4594 mol$

Mass of solvent = 186 g = 0.186 kg

$m=\frac{0.4594 mol}{0.186 kg}=2.4610 mol/kg$

The molality of methanol is 2.4610 mol/kg.