Question

The solubility of CO2 in water exposed to the atmosphere, where the partial pressure of CO2 is 3.87 10 atm, is 1.32 * 10 M. At the same temperature, what would be the solubility of pressurized CO2, at a pressure of 68 atm? a. 5.09 x 10*M b. 2.32 M C. 1920 M d. 4310 M How many mL of a 0.575 M glucose solution (MW=184 g/mol) would be needed to make 350 ml of a 0.320 M glucose solution? a. 631 b. 195 c. 67 d. 0.21

Answer 1

How many ml of a 0.575M glucose solution would be needed to make 350ml of a 0.320M glucose solution?

Use the formula :

M₁V₁ = M₂ V₂ --------------------------------(1)

Where M1 = Initial molarity = 0.575M

V₁ = initial volume = ?

M₂ = final molarity = 0.320M

V₂ = final volume = 350ml

Substitute the given values in equation (1), we get

0.575M x V₁ = 0.320M x 350ml

=> V₁ = 0.320M x 350ml / 0.575M

=> V₁ = 194.7 mL

= 195 mL

option b. 195 is the answer.

This question is based on Henry's law:

According to this law, the solubility of a gas in a liquid at constant temperature is directly proportional to the partial pressure of the gas which is in equilibrium with the liquid.

S $\alpha$ p

=> s/p = constant

The constant is called Henry's law constant and every gas has its Henry's law constant value.

where s=solubility of the gas

p = partial pressure of the gas

Let the solubility of the gas initially be s₁

Initial partial pressure be p₁

final solubility be s₂

and final partial pressure be p₂

S₁ / P₁ = CONSTANT

S₂ /P₂ = CONSTANT

then, according to Henry's law:

s₁ / p₁ = s₂ / p₂ ----------------------------------(i)

Let the solubility of the gas initially be s₁ and it is given 1.32x10^-5M

Initial partial pressure be p₁ = 3.87x10^-4 atm

final solubility be s₂ = ?

and final partial pressure be p₂ = 68 atm

Substitute this values in equation (i), we get

1.32x10^-5M / 3.87x10^-4 atm = s₂ / 68 atm

=> s₂ = [1.32x10^-5M / 3.87x10^-4 atm ] x 68 atm

=> s₂ = 2.32M

Hence, the solubility of CO₂ at 68 atm of partial pressure is 2.32 M

option b. 2.32M is the answer.