Question

Calculate the concentration of CO ₂ in orange soda that
was bottled (at a certain temperature) with a partial pressure of CO ₂  of 3.15 atm over the liquid.   
At that temperature, the Henry's law constant, k, for CO ₂ is 3.05 x 10 ^-2
mol/L atm.
(i.e., 3.05 x 10 ^-2
  M/atm)

		14.0   M
		8.12 M
		2.74  M
		6.63 x 10 -3  M
		0.0961 M
		23.8 M
		11.9   M
		7.93 x 10 -1   M

2.

What is the mole fraction of sucrose when 433 g sucrose C ₁₂H ₂₂O ₁₁ (molar mass 342.3 g/mol) is dissolved in 655 g water?

		0.0336
		0.101
		0.0456
		1.03
		1.75

3.

A bottle of orange beverage has CO ₂ concentration equal to 0.0801 Molar at a certain temperature. What is the partial pressure of CO ₂ in the container above the liquid in atm?  
At this temperature, the Henry's law constant, k, for CO ₂ is 3.60 x 10 ^-2
mol/L atm. (i.e., 3.20 x 10 ^-2   M/atm)

		0.842 atm
		5.28 atm
		1.55 x 10-1 atm
		3.20 x 10-3 atm
		0.422 atm
		2.23 atm
		4.51 atm
		2.98 atm

Answer 1

Answer:-

(1)-

Given:-

partial pressure of CO2 (p) = 3.15 atm

Henry's law constant (k) for CO2 = 3.05 10-2 mol / L atm = 3.05 10-2 M / atm

concentration of CO2 (c) = ?

As we know that according to the Henry's Law , the mass of a gas dissolved per unit volume of a solvent i.e concentration (c) is directly proportional to the partial pressure (p) of the gas in equilibrium with the solution at a given temperature.

So

concentration of a gas (c)   partial pressure (p) of the gas

then

concentration of a gas (c) =  Henry's law constant (k) partial pressure (p) of the gas

therefore

concentration of a CO2 gas (c) =  Henry's law constant (k) of CO2 gas partial pressure (p) of the CO2 gas

concentration of a CO2 gas (c) =  3.05 10-2 M / atm   3.15 atm

concentration of a CO2 gas (c) =  9.61 ​​​​​​​10-2 M

concentration of a CO2 gas (c) = 0.0961 M (i.e the answer)

(2)-

Given:-

wt. of sucrose (C12H22O11) i.e (wsucrose) = 433 g

molar mass of sucrose (C12H22O11) i.e (Msucrose) = 342.3 g/mol

wt. of water (H2O) i.e (wwater) = 655 g

Since we know that

molar mass of water (H2O) i.e (Mwater) = 18 g/mol

As we know that

no. of moles of compound (ncompound) = wt. of compound (wcompound) / molar mass of compound (Mcompound)

therefore

no. of moles of compound (nsucrose) = wt. of compound (wsucrose) / molar mass of compound (Msucrose)

no. of moles of compound (nsucrose) = 433 g / 342.3 g/mol

no. of moles of compound (nsucrose) = 1.26 mol

similarly

no. of moles of compound (nwater) = wt. of compound (wwater) / molar mass of compound (Mwater)

no. of moles of compound (nwater) = 655 g / 18 g/mol

no. of moles of compound (nwater) = 36.39 mol

Also we know that

total no. of moles in solution (ntotal) = no. of moles of compound (nsucrose) + no. of moles of compound (nwater)

total no. of moles in solution (ntotal) = 1.26 mol + 36.39 mol

total no. of moles in solution (ntotal) = 37.65 mol

So according to the formula

mole fraction of sucrose (C12H22O11) i.e (sucrose) = no. of moles of compound (nsucrose) / total no. of moles in solution (ntotal)

mole fraction of sucrose (C12H22O11) i.e (sucrose) = 1.26 mol / 37.65 mol

mole fraction of sucrose (C12H22O11) i.e (sucrose) = 0.0335 (i.e the answer)

(3)-

Given:-

Henry's law constant (k) for CO2 = 3.60 10-2 mol / L atm = 3.60 ​​​​​​​10-2 M / atm

concentration of CO2 (c) = 0.0801 M

partial pressure of CO2 (p) = ?

As we know that

concentration of a gas (c) =  Henry's law constant (k) partial pressure (p) of the gas

therefore

0.0801 M =  3.60 ​​​​​​​10-2 M / atm partial pressure (p) of the CO2 gas

partial pressure (p) of the CO2 gas = 0.0801 M / 3.60 ​​​​​​​10-2 M / atm

partial pressure (p) of the CO2 gas = 0.0801​​​​​​​102 atm / 3.60

partial pressure (p) of the CO2 gas = 0.0223 ​​​​​​​ 102 atm

partial pressure (p) of the CO2 gas = 0.0223 ​​​​​​​ 100 atm

partial pressure (p) of the CO2 gas = 2.23 atm (i.e the answer)