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Olenka [21]
2 years ago
10

You have to prepare some 2 M solutions, with 10 g of solute in each. What volume of solution will you prepare, for each solute b

elow? (Ar:H51,Li57,N514,O516,Mg524,S532.)
a lithium sulfate, Li2SO4
Chemistry
1 answer:
alexdok [17]2 years ago
8 0

Answer:

             0.045 L  or 45 mL

Explanation:

Moles = Mass/M.Mass

Moles = 10 g / 109.94 g/mol

Moles = 0.09 moles

Also,

Molarity = Moles / Vol in L

Or,

Vol in L = Moles / Molarity

Vol in L = 0.09 mol / 2 mol/L

Vol in L = 0.045 L

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Nitrogen effuses through a pinhole 1.7 times as fast as another gaseous element under the same conditions. Estimate the other el
marysya [2.9K]

Answer:

80.92, Krypton

Explanation:

<u>What is effusion?</u>

• It is a process where gas escapes through a pinhole (a very small hole) into a region of low pressure or vacuum

<u>Graham's law of effusion of </u><u>gas</u>

• states that at a given constant temperature and pressure, the rate of effusion of gases is inversely proportional to the square root of their molar masses

\boxed{ \frac{Rate_1}{Rate_2} =  \sqrt{ \frac{M_2}{M_1} } }

<u>Calculations</u>

Nitrogen exist as N₂ at room temperature, thus its molar mass is 2(14)= 28.

Let the rate and molar mass of unknown gas be Rate₂ and M₂ respectively.

Since N₂ effuses 1.7 times as fast as the unknown gas,

Rate₁= 1.7(Rate₂)

\frac{Rate_1}{Rate_2} = 1.7

1. 7 =  \sqrt{ \frac{M_2}{28} }

Square both sides:

2.89  = \frac{M_2}{28}

Multiply both sides by 28:

2.89(28)= M₂

M₂= 80.92

<u>Identity of </u><u>gas</u>

The molar mass of 80.92 lies between Bromine and Krypton. However since Bromine exist as Br₂, the value of it's molar mass would be 159.8 instead. Hence, Bromine is eliminated.

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Thus based on the above reasoning, its probable identity is Krypton.

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