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A 1.19 g sample of an unknown halogen occupies 109 mL at 398 K and 1.41 atm. What is the identity of the halogen

Sagot :

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Answer:

The halogen is Iodine.

Explanation:

Using the ideal gas equation, we find the number of moles of gas present, n.

PV = nRT where P = pressure of gas = 1.41 atm, V = volume of gas = 109 mL = 0.109 L, n = number of moles of gas, R = molar gas constant = 0.082 L-atm/mol-K and T = temperature of gas = 398 K

Since PV = nRT, making n subject of the formula, we have

n = PV/RT

substituting the values of the variables into the equation, we have

n = 1.41 atm × 0.109 L/(0.082 L-atm/mol-K × 398 K)

n = 0.15369 atm-L/32.636 L-atm/mol

n = 0.0047 mol

Since n = m/M where m = mass of gas = 1.19 g and M = relative molecular mass of gas

So, M = m/n

M = 1.19 g/0.0047 mol

M = 252.7 g

Since halogens are diatomic the relative atomic mass is M/2 = 252.7g/2 = 126.34 g

From tables, the only halogen with this atomic mass is Iodine.

So, the halogen is Iodine.

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