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3 years ago

What is the concentration in mol L-l of a 12% solution of tetrahydrate magnesium chloride (MgCl2)?

Chemistry

1 answer:

madreJ [45]3 years ago

3 0

The concentration in mol L-l (M) = 0.717

<h3>Further explanation</h3>

Given

12% solution of tetrahydrate magnesium chloride (MgCl₂)

Required

The concentration

Solution

Tetrahydrate magnesium chloride (MgCl₂)⇒MgCl₂.4H₂O

MW = Ar Mg+2. Ar Cl+8. Ar H + 4. Ar O

MW=24.31 + 2 x 35.45 + 8 x 1.01 + 4 x 16

MW=24.31+70.9+8.08+64

MW=167.29 g/mol

12% solution = 12 % m/v = 12 g in 100 ml solution

For 1 L solution :

$\tt \dfrac{1}{0.1}\times 12~g=120~g$

The concentration in g/L = 120 g/L

Convert grams to moles :

$\tt mol=\dfrac{120}{167.29}=0.717$

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How many liters of hydrogen gas will be produced at STP from the reaction of 7.179×10^23 atoms of magnesium with 54.219g of phos

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Answer: The volume of hydrogen gas produced will be, 12.4 L

Explanation : Given,

Mass of $H_3PO_4$ = 54.219 g

Number of atoms of $Mg$ = $7.179\times 10^{23}$

Molar mass of $H_3PO_4$ = 98 g/mol

First we have to calculate the moles of $H_3PO_4$ and $Mg$ .

$\text{Moles of }H_3PO_4=\frac{\text{Given mass }H_3PO_4}{\text{Molar mass }H_3PO_4}$

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and,

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Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

$3Mg+2H_3PO_4\rightarrow Mg(PO_4)_2+3H_2$

From the balanced reaction we conclude that

As, 3 mole of $Mg$ react with 2 mole of $H_3PO_4$

So, 0.553 moles of $Mg$ react with $\frac{2}{3}\times 0.553=0.369$ moles of $H_3PO_4$

From this we conclude that, $H_3PO_4$ is an excess reagent because the given moles are greater than the required moles and $Mg$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $H_2$

From the reaction, we conclude that

As, 3 mole of $Mg$ react to give 3 mole of $H_2$

So, 0.553 mole of $Mg$ react to give 0.553 mole of $H_2$

Now we have to calculate the volume of $H_2$ gas at STP.

As we know that, 1 mole of substance occupies 22.4 L volume of gas.

As, 1 mole of hydrogen gas occupies 22.4 L volume of hydrogen gas

So, 0.553 mole of hydrogen gas occupies $0.553\times 22.4=12.4L$ volume of hydrogen gas

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3 years ago

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Answer:

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b) 5.62 MeV / nucleon

c) 8.80 MeV / nucleon

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Explanation:

Binding energy = ( Δm * 931.5 ) MeV

Binding energy per nucleon = Binding energy in / Number of nucleon

<u>a) ²H = 1 neutron , 1 proton = 2 nucleons </u>

Given that the theoretical mass = 2.0141 u

Actual mass = 1.0078 u + 1.0087 u = 2.0165 u

Δm = 2.0165 u - 2.0141 u = 2.4 * 10^-3 u

∴ Binding energy per nucleon = ( 2.4 * 10^-3 * 931.5 ) MeV / 2 nucleons

= 1.12 MeV / nucleon

<u>b) ⁷Li = 3 protons , 4 neutrons = 7 nucleons </u>

theoretical mass = 7.0160 u

Actual mass = ( 3 * 1.0078 ) + ( 4 * 1.0087 ) = 7.0582 u

Δm = ( 7.0582 u - 7.0160 u ) = 0.0422 u

∴ Binding energy per nucleon = ( 0.0422 * 931.5 ) / 7

= 5.62 MeV / nucleon

<u>C) ⁶²Ni = 28 protons , 34 neutrons = 62 nucleons </u>

Theoretical mass = 61.9283 u

Actual mass = ( 28 * 1.0078 ) u + ( 34 * 1.0087 ) u

= 62.5142 u

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∴ Binding energy per nucleon = ( 0.5859 * 931.5 ) / 62

= 8.80 MeV / nucleon

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= 110.9138 u

Δm = ( 110.9138 - 109.9030 ) = 1.0108 u

∴ Binding energy per nucleon = ( 1.0108 * 931.5 ) / 110

= 8.56 MeV / nucleon

hence we can conclude that the binding energy has a maximum value for nuclei with a mass around 60

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3 years ago