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Leona [35]
3 years ago
12

What is the molar mass of a gas which has a density of 0.00249 g/mL at 20.0 degrees celcius and 744.0 mm Hg?

Chemistry
1 answer:
creativ13 [48]3 years ago
8 0

61.24 is the molar mass of a gas which has a density of 0.00249 g/mL at 20.0 degrees celcius and 744.0 mm Hg.

Explanation:

given that:

density = 0.00249 g/ml (\frac{mass}{volume})  or 2.49 grams/litre

P = 744 mm Hg OR 0.978 atm

T = 20 Degrees or 293.15 Kelvin

R = 0.08206 Litre atm/mole K

molar mass =?

Formula used/

PV = nRT    equation 1

here n is number of moles:

n = \frac{mass}{atomic mass}

putting the value of n and value of density in the equation 1:

PV = \frac{mass}{molar mass} x RT

molar mass = \frac{mass}{volume} x \frac{RT}{P}

                    = density x \frac{RT}{P}

                     = \frac{2.49 x 0.08276 x 293.15}{0.978}

                     = 61.24 is the molar mass of the gas.

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Calculate the freezing point of a solution made from 220g of octane (C Hua), molar mass = 114,0 gmol dissolved in 1480 g of benz
stiv31 [10]

Answer: Freezing point of a solution will be -1.16^0C

Explanation:

Depression in freezing point is given by:

\Delta T_f=i\times K_f\times m

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i= vant hoff factor = 1 (for non electrolyte)

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m= molality

\Delta T_f=i\times K_f\times \frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}

Weight of solvent (benzene)= 1480 g =1.48 kg

Molar mass of solute (octane) = 114.0 g/mol

Mass of solute (octane) = 220 g

(5.50-T_f)^0C=1\times 5.12\times \frac{220g}{114.0 g/mol\times 1.48kg}

(5.50-T_f)^0C=6.68

T_f=-1.16^0C

Thus the freezing point of a solution will be -1.16^0C

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It is a molecular compound, which is a mixture of at least two atoms—the smallest building blocks of matter—joined by a covalent bond. These atoms are joined by a covalent bond, which is formed when electrons are shared.

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