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aleksandr82 [10.1K]
3 years ago
7

A 13.5 g sample of an unknown gas occupies 5.10 L at 149.83 kPa and 301 K. What is the molar mass of the gas ?

Chemistry
1 answer:
alisha [4.7K]3 years ago
7 0

Answer:

The molar mass of the gas is 44.19 g/mol

Explanation:

Amount of sample of gas = m = 13.5 g

Volume occupied by the gas = V = 5.10 L

Pressure of the gas = P = 149.83 KPa

1 KPa = 0.00986 atm

P = 149.83 \textrm{ KPa} \times 0.00986 \textrm{ atm/KPa} = 1.48 \textrm{ atm}

Assuming M g/mol to be the molar mass of the gas

Assuming the gas is behaving as an ideal gas

\textrm{PV} =\textrm{nRT} \\\textrm{PV} = \displaystyle \frac{m}{M}\textrm{ RT } \\1.48 \textrm{ atm}\times 5.10 \textrm{ L} = \displaystyle \frac{13.5 \textrm{ g}}{M}\times 0.0821 \textrm{ L.atm.mol}^{-1}.K^{-1}\times 301\textrm{K} \\M = 44.19 \textrm{ g/mol}

The molar mass of gas is 44.19 g/mol

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The compound dioxane, which is used as a solvent in various industrial processes is composed of C,H, and O atoms. Combustion of
Lady_Fox [76]

Answer:

The correct formula of dioxane is C₄H₈O₂.

Explanation:

Given data:

mass of dioxane= 2.23 g

mass of water = 1.802 g

mass of carbon dioxide = 4.401 g

molar mass of dioxane = 88.1 g / mol

Molecular formula of dioxane = ?

Solution:

percentage of carbon = (4.401 g/2.23 g ) × (12 /44) × 100

                                     =  (1.98 × 0.273) × 100 = 54.1

percentage of hydrogen =  (1.802 g/ 2.23 g) × (2.016 /18.016) × 100

                                         = (0.81 × 0.112) × 100 = 9.072

percentage of oxygen = 100 - (54.1 + 9.072)

                                     = 100 - 63.172 = 36.828

Now we will determine the number of grams atoms of carbon, hydrogen and oxygen.

No. of gram atoms of carbon = 54.1 /12 = 4.51

No. of gram atoms of hydrogen = 9.072 / 1.008 = 9

No. of gram atoms of oxygen = 36.828 / 16 = 2.302

Atomic ratio:

    C :H :O               4.51/ 2.302    :   9 / 2.302   :  2.302 /2.302

    C :H :O                2 : 4 : 1

Molecular formula:

   Molecular formula = n × (empirical formula)

   n = molar mass of compound / empirical formula mass

   empirical formula mass= 2 × 12 + 4 × 1.008  + 1 × 16

    empirical formula mass= 24+ 4.032 +16 = 44.032

                 n = 88.1 / 44.032 = 2

        Molecular formula = n × (empirical formula)

        Molecular formula = 2 × (C₂H₄O)

         Molecular formula = C₄H₈O₂

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