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ELEN [110]
2 years ago
15

a student used 10 mL water instead of 30 mL for extraction of salt from mixture. How may this change the percentage of NaCl extr

acted?
Chemistry
1 answer:
Anit [1.1K]2 years ago
5 0
It will be extracted only 1/3 of NaCl less in 10 mL of water than in 30 mL of water.

If it is known that solubility of NaCl is 360 g/L, let's find out how many NaCl is in 30 mL of water:

360 g : 1 L = x g : 30 mL

Since 1 L = 1,000 mL, then:
360 g : 1,000 mL = <span>x g : 30 mL

Now, crossing the products:
x </span>· 1,000 mL = 360 g · 30 mL
x · 1,000 mL = 10,800 g mL
x = 10,800 g ÷ 1,000 
x = 10.8 g

So, from 30 mL mixture, 10.8 g of NaCl could be extracted.

Let's calculate the same for 10 mL water instead of 30 mL.

360 g : 1 L = x g : 10 mL

Since 1 L = 1,000 mL, then:
360 g : 1,000 mL = <span>x g : 10 mL

Now, crossing the products:
x </span>· 1,000 mL = 360 g · 10 mL
x · 1,000 mL = 3,600 g mL
x = 3,600 g ÷ 1,000 
<span>x = 3.6 g
</span>
<span>So, from 10 mL mixture, 3.6 g of NaCl could be extracted.
</span>
Now, let's compare:
If from 30 mL mixture, 10.8 g of NaCl could be extracted and <span>from 10 mL mixture, 3.6 g of NaCl could be extracted, the ratio is:
</span>3.6/10.8 = 1/3

Therefore, i<span>t will be extracted only 1/3 of NaCl less in 10 mL of water than in 30 mL of water. </span>
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Step 1:  write  the balanced equation between CH3COOH  + NaOH

that is CH3COOH   + NaOH  →  CH3COONa  + H2O


step 2 :  find the moles of NaOH

moles  =molarity  x volume in L

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moles = 0.175 mol /L x 0.0269 L  =0.0047  moles  of NaOH


Step 3: use the mole  ratio to find moles of CH3COOH

that is the  mole ratio of  CH3COOH: NaOH is 1:1 therefore  the moles of CH3COOH is  =0.0047  moles


Step 4:  find the  molarity  of  CH3COOH

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A filament of a light bulb is made from a pure sample of tungsten.
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The correct options are:

1) Pure chemical substance

2) Element.

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each of the term is explained below:

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2) False by definition of mixture.

3) false by definition of mixture.

4) <u>Solution:</u> Tungsten in a filament is in solid form hence solution is incorrect.

5) Since it is given in the question itself that the sample is pure hence option 'E' is correct as chemical substance is a pure substance that has a homogeneous composition and the sample in the question is given as pure.

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<u>Answer:</u> The empirical and molecular formula for the given organic compound is CH_3O and C_4H_{12}O_4

<u>Explanation:</u>

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

C_xH_yO_z+O_2\rightarrow CO_2+H_2O

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of CO_2=9.06g

Mass of H_2O=5.58g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

  • <u>For calculating the mass of carbon:</u>

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 9.06 g of carbon dioxide, \frac{12}{44}\times 9.06=2.47g of carbon will be contained.

  • <u>For calculating the mass of hydrogen:</u>

In 18g of water, 2 g of hydrogen is contained.

So, in 5.58 g of water, \frac{2}{18}\times 5.58=0.62g of hydrogen will be contained.

  • Mass of oxygen in the compound = (6.38) - (2.47 + 0.62) = 3.29 g

To formulate the empirical formula, we need to follow some steps:

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{2.47g}{12g/mole}=0.206moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.62g}{1g/mole}=0.62moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{3.29g}{16g/mole}=0.206moles

  • <u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.206 moles.

For Carbon = \frac{0.206}{0.206}=1

For Hydrogen  = \frac{0.62}{0.206}=3

For Oxygen  = \frac{0.206}{0.206}=1

  • <u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : H : O = 1 : 3 : 1

Hence, the empirical formula for the given compound is C_1H_{3}O_1=CH_3O

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is:

n=\frac{\text{molecular mass}}{\text{empirical mass}}

We are given:

Mass of molecular formula = 124 amu = 124 g/mol

Mass of empirical formula = 31 g/mol

Putting values in above equation, we get:

n=\frac{124g/mol}{31g/mol}=4

Multiplying this valency by the subscript of every element of empirical formula, we get:

C_{(1\times 4)}H_{(3\times 4)}O_{(1\times 4)}=C_4H_{12}O_4

Thus, the empirical and molecular formula for the given organic compound is CH_3O and C_4H_{12}O_4

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