Suppose 0.510 g of a gas occupies a volume of 0.175 l at a temperature of 25.0 oc is held at a pressure of 1.000 atm. what is the molar mass of the gas? express your answer in g/mol.

2b)a 1.00 l sample of dry air contains 0.0319 mol n2, 0.00856 mol o2, and 0.000381 mol ar. if temperature is 25.0 oc what is the partial pressure of n2? express your answer in atmospheres.

1) Molar Mass = 71.3g/mol

2) Partial Pressure of N2 = 0.78031 atm

Explanation:

First Question;

Mass (m) = 0.510g

Volume (V) = 0.175L

Temeperature (T) = 25 C + 273 = 298K (Converting to Kelvin Temperature)

Pressure (p) = 1 atm

The formular relationship between mass and Molar mass is given as;

Number of moles (n) = mass (m) / molar mass (M)

Using Ideal gas equation, we would be able to calculate the number of moles;

PV = nRT; where R = gas constant = 0.0821 L atm K−1 mol−1

n = PV / RT

n = (1 * 0.175) / (0.0821 * 298)

n = 0.175 / 24.4658

n = 0.007153 moles

Substituting the value of n in Number of moles (n) = mass (m) / molar mass (M)

Molar mass = mass / Number of moles

Molar mass = 0.510 / 0.007153 = 71.3g/mol

Second Question;

Key terms;

mole fraction: number of moles of one particular gas divided by the total moles of gas in the mixture

Total moles of gas = 0.0319 mol N2 + 0.00856 mol O2 + 0.000381 mol Ar

Total moles of gas = 0.040841 moles

Mole fraction of N2 = moles of N2 / Total moles = 0.0319 / 0.040841 = 0.7811

Volume = 1L

Temperature = 25C + 273 = 298k (converting to Kelvin Temperature)

Number of moles = 0.040841

Total pressure = ?

Using Ideal gas equation;

PV = nRT; where R = gas constant = 0.0821 L atm K−1 mol−1

P = nRT / V

P = (0.040841 * 0.0821 * 298) / 1

Total Pressure = 0.999 atm

The partial pressure of one individual gas within the overall mixtures, pi, can be expressed as follows:

Pi=Ptotal * xi

where xi is the mole fraction.

Partial Pressure of N2 = 0.999 * 0.7811

Partial Pressure of N2 = 0.78031 atm

20 drops hope this

answer:

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