Avolume of 18.0 l contains a mixture of 0.250 mole n2 , 0.250 mole o2 , and an unknown quantity of he. the temperature of the mixture is 0 ∘c , and the total pressure is 1.00 atm . how many grams of helium are present in the gas mixture?

- 1.22 g of helium are present in gas mixture.

Solution:- Given, total pressure = 1.00 atm

T = 0 + 273 = 273 K

V = 18.0 L

From ideal gas law equation, PV = nRT

P = nRT/V

Partial pressure of nitrogen = (0.250 x 0.0821 x 273)/18.0

partial pressure of nitrogen = 0.311 atm

similarly, partial pressure of oxygen = (0.250 x 0.0821 x 273)/18.0

partial pressure of oxygen = 0.311 atm

From Dalton's law of partial pressure, total pressure is the sum of partial pressures of the gases present in the container.

So, total pressure = (partial pressure of nitrogen + partial pressure of oxygen + partial pressure of helium)

partial pressure of helium = 1.00 atm - (0.311 atm + 0.311 atm)

partial pressure of helium = 1.00 atm - 0.622 atm

partial pressure of helium = 0.378 atm

n = pV/RT

So, moles of helium = (0.378 x 18.0)/(0.0821 x 273) = 0.304 mol

Now we can calculate the grams of helium on multiplying it's moles by it's molar mass.

0.304 mol x (4.00g/1mol) = 1.22 g

answer: option (b) is the correct answer.

explanation:

a covalent bond is formed by sharing of electrons. whereas polarity arises due to difference in electronegativity of two atoms.

therefore, more is the electronegativity of an atom which combines with the titanium, more readily it will induce a partial positive and negative charge.

thus, out of the given options fluorine is most electronegative and has an incomplete octet. therefore, it will readily combine with titanium to share the electrons and complete its octet.