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boyakko [2]
2 years ago
13

Calculate the molar mass of a gas at 388 torr and 45°c if 206 ng of gas occupies 0.206 ul. (

Chemistry
1 answer:
Dmitrij [34]2 years ago
6 0
To make our solution more systematic, let's convert all units that is consistent with the units of R which is 0.0821 L-atm/mol-K.

For pressure: 760 torr = 1 atm
388 torr * 1 atm/760 torr = 0.5105 atm

For volume, 1 μL = 10⁻⁶ L
0.206 μL * 10⁻⁶ L/1 μL = 2.06×10⁻⁷ L

For temperature,
T = 45 + 273 = 318 K

For mass, 1 ng = 10⁻⁹ g
206 ng * 10⁻⁹ g/1 ng = 2.06×10⁻⁷ g

Assuming ideal gas,

PV=nRT
(0.5105 atm)(2.06×10⁻⁷ L) = n(0.0821 L-atm/mol-K)(318 K)
n = 4×10⁻⁹ mol

Molar mass = Mass/n = 2.06×10⁻⁷ g/4×10⁻⁹ mol
<em>Molar mass = 51.14 g/mol</em>

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Lithium acetate, LiCH3CO2, is a salt formed from the neutralization of the weak acid acetic acid, CH3CO2H, with the strong base
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Answer : The pH of 0.289 M solution of lithium acetate at 25^oC is 9.1

Explanation :

First we have to calculate the value of K_b.

As we know that,

K_a\times K_b=K_w

where,

K_a = dissociation constant of an acid = 1.8\times 10^{-5}

K_b = dissociation constant of a base = ?

K_w = dissociation constant of water = 1\times 10^{-14}

Now put all the given values in the above expression, we get the dissociation constant of a base.

1.8\times 10^{-5}\times K_b=1\times 10^{-14}

K_b=5.5\times 10^{-10}

Now we have to calculate the concentration of hydroxide ion.

Formula used :

[OH^-]=(K_b\times C)^{\frac{1}{2}}

where,

C is the concentration of solution.

Now put all the given values in this formula, we get:

[OH^-]=(5.5\times 10^{-10}\times 0.289)^{\frac{1}{2}}

[OH^-]=1.3\times 10^{-5}M

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pOH=-\log [OH^-]

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pOH=4.9

Now we have to calculate the pH.

pH+pOH=14\\\\pH=14-pOH\\\\pH=14-4.9=9.1

Therefore, the pH of 0.289 M solution of lithium acetate at 25^oC is 9.1

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2 years ago
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