For a sample of acetylene collected over water, total gas pressure is 760 torr and the volume is 459 ml. at the temperature of the gas (23°c), the vapor pressure of water is 21.0 torr. how many grams of acetylene are collected?

0.52g

Explanation:

To calculate the mass of acetylene collected, we can calculate the number of moles of acetylene collected and multiply this by the molar mass of acetylene.

To calculate the number of moles of acetylene collected, we can use the ideal gas equation I.e PV = nRT

Rearranging the equation, n =PV/RT

We now identify each of the terms below before substituting and calculating.

n = number of moles, which we are calculating.

R = molar gas constant = 62.64 L.Torr. K^-1. mol^-1

V = volume = 459ml : 1000ml ÷ 1L, hence , 459ml = 459/1000 = 0.459L

T = temperature = 23 degrees Celsius = 273 + 23 = 296K

P = pressure. But since the gas was collected over water, we subtract the vapour pressure of water from the total pressure = 760 - 21 = 739torr

We substitute these values into the equation to yield the following:

n = (739 × 0.459) ÷ ( 62.64 × 300)

n = apprx 0.02 moles

To calculate the mass of acetylene collected, we need the molar mass of acetylene. The molecular formula of acetylene = C2H2, atomic mass of carbon = 12 and atomic mass of hydrogen = 1, thus , the molar mass = 2(12) + 2(1) = 26g/mol

Thus the mass of acetylene collected = 0.02 mole × 26g/mol = 0.52g

43.452

step-by-step explanation:

284 times .153 = 43.452