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calculate the molarity of MgCl2 in the following solution: 5.34 g of MgCl2 is dissolved and diluted to 214 mL of solution.

anyanavicka [17]

$\quad\hookrightarrow\quad \sf {0.262M }$

<h3>Ⲋⲟⳑⳙⲧⳕⲟⲛ :</h3>

Molarity is used to measure the concentration of a solution , so it is also as molar concentration. It is denoted as M or Mol/L

We are given that :

Weight of $\sf MgCl_{2}$ = 5.34g
Volume of solution = 214 ml , or 0.214 L

The molar mass of magnesium chloride ( $\sf MgCl_{2}$ ) is 95.21 g / mol

We can calculate the molarity of the solution by dividing the number of moles of solute by volume of solvent in liter ,i.e:

$\quad\longrightarrow\quad \sf {M = \dfrac{n}{V} }$ ‎ㅤ‎ㅤ‎ㅤ⸻( 1 )

Where,

M = molarity
n = number of moles
V = Volume

We can calculate the number of moles by dividing the actual mass by its molar mass ,i.e:

$\quad\longrightarrow\quad \sf { n = \dfrac{w}{m}}$ ‎ㅤ‎ㅤ‎ㅤ‎⸻ ( 2 )

Where,

n = number of moles
m = molar mass
w = actual mass

Therefore,

$\implies\quad \tt {n =\dfrac{w}{m} }$

$\implies\quad \tt { n =\dfrac{5.35\: g}{95.21\: g /mol}}$

$\implies\quad{\pmb{ \tt {n = 0.056 mol}} }$

Putting the values in equation ( 1 ):

$\implies\quad \tt {M=\dfrac{n}{V} }$

$\implies\quad \tt { M =\dfrac{0.056\:mol}{0.214\:L}}$

$\implies\quad\underline{\pmb{ \tt { M = 0.262 \:M }}}$

7 0

2 years ago

A nitrogen oxide, containing 53.85% N, acts as a vasodilator, lowering blood pressure in the human body. What is its empirical f

jenyasd209 [6]

The empirical formula of the nitrogen oxide that acts as a vasodilator and lowers human blood pressure is $N_4O_3$ . Empirical formula is the smallest whole number ratios of elements in a compound.

FURTHER EXPLANATION

To get the empirical formula from mass percent data, the following steps are followed:

1. Get the mass percent data for all the elements that make up the compound.

2. Assume that there is 100 grams of sample compound. Determine the equivalent masses of each element using the mass percent given.

3. Convert the mass of each element to number of moles.

4. Divide each calculated mole by the smallest mole value.

5. The quotient will be the subscript of the element in the compound. If a decimal is obtained, round off the answers to the nearest whole number. Some exceptions to this are the following decimals: x.25, x.33, x.50, and x.75. When these decimals are obtained, all the subscripts are multiplied by a number that will result in a whole number.

Applying the steps to the problem,

STEP 1: Get the mass percent.

Mass % of N = 53.85

Mass % of O = (100 - 53.85) = 46.15

STEP 2: Assume that 100 g sample is used and convert the mass % to mass in grams.

mass of N = 53.85% x 100 g = 53.85 g

mass of O= 46.15% x 100 g = 46.15 g

STEP 3: Convert mass to moles by dividing the given mass by the formula mass.

$moles \ of \ N \ = 53.85 \ g (\frac{1 \ mol}{14 \ g}) = 3.85 \\\\moles \ of \ O \ = 46.26 \ g \ (\frac{1 \ mol}{16 \ g}) \ = 2.89$

The moles may be written as the temporary subscripts of the elements in the compound as follows:

$N_{3.85}O_{2.89}$

STEP 4: Divide the moles by the smallest mole value.

$N_{3.85}O_{2.89}\\N_{ \frac{3.85}{2.89}} \ O _{ \frac{2.89}{2.89}} \\ N_{1.33}O_{1}\\$

STEP 5: Multiply the subscripts by a number that would give the smallest whole number ratio.

Since the decimal is 1.33 it cannot be rounded off to 1. It should be multiplied by 3 to get the smallest whole number.

NOTE: All subscripts must be multiplied by the same factor, 3.

$N_{1.33}O_{1}\\3(N_{1.33}O_{1})\\\boxed {N_{4}O_{3}}$

To check if the empirical formula is correct, calculate the mass % of each element based on the formula and compare with the given in the problem.

$mass \ \%\ = (\frac{mass \ of \ element}{mass \ of \ compound}) x 100\\\\mass \ \% \ of \ N = \ \frac{(4)(14)}{(4)(14) \ + \ (3)(16)}$

$mass \ \% \ of \ N \ = 53.85%\\mass \ \% \ of \ O \ = (\frac{(3)(16)}{(4)(14)+(3)(16)}) \ 100\\mass \ \% \ of \ O \ = 46.15%$

Since the values are the same from the given, the answer is correct.

LEARN MORE

Writing Chemical Formula brainly.com/question/4697698
Naming Compounds brainly.com/question/8968140
Mole Conversion brainly.com/question/12979299

Keywords: empirical formula, compounds

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