The solubility of nitrogen gas at 25 degrees c and 1 atm is 6.8 x 10^(-4) mol/l. if the partial pressure of nitrogen gas in air is 0.76 atm, what is the concentration (molarity) of dissolved nitrogen? 5.2 x 10^(-4) m 1.1 x 10^(-5) m 4.9 x 10^(-4) m 3.8 x 10^(-4) m 6.8 x 10^(-4) m

5.2 x 10⁻⁴ M.

Explanation:

P = kC

where P is the partial pressure of the gaseous  solute above the solution.

k is a constant (Henry’s constant).

C is the concentration of the dissolved gas.

P₁C₂ = P₂C₁,

P₁ = 1.0 atm, C₁ = 6.8 x 10⁻⁴ mol/L.

P₂ = 0.76 atm, C₂ = ??? mol/L.

∴ C₂ = (P₂C₁)/P₁ = (0.76 atm)(6.8 x 10⁻⁴ mol/L)/(1.0 atm) = 5.168 x 10⁻⁴ mol/L ≅ 5.2 x 10⁻⁴ M.

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g