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3 years ago

11

To interconvert the concentration units molality (m) and mass percent, you must also know the density of the solution. A) True B

) False

1 answer:

uranmaximum [27]3 years ago

7 0

Answer:

The correct answer is option false.

Explanation:

Molality of the solution defined as moles of substance present in 1 kilogram of solvent.

Moles = $\frac{Mass of solute}}{\text{molar mass of solute}}$

$Molality=\frac{Moles}{\text{Mass of solvent(kg)}}$

Mass of percent (w/w%) of the solution is defined as amount of solute present in 100 grams of solution.

$(w/w\%)_=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 100$

So, if want to inter-convert molality into mass percent we can do that without knowing density of solution.

Mass of solution = Mass of solute + Mass of solvent

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Enter your answer in the box provided. How many grams of helium must be added to a balloon containing 6.24 g helium gas to doubl

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Answer : The mass of helium gas added must be 12.48 grams.

Explanation : Given,

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Molar mass of helium = 4 g/mole

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$\text{Moles of }He=\frac{\text{Mass of }He}{\text{Molar mass of }He}=\frac{6.24g}{4g/mole}=1.56moles$

Now we have to calculate the moles of helium gas at doubled volume.

According to the Avogadro's law, the volume of gas is directly proportional to the number of moles of gas at same pressure and temperature. That means,

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or,

$\frac{V_1}{V_2}=\frac{n_1}{n_2}$

where,

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$V_2$ = final volume of gas = 2V

$n_1$ = initial moles of gas = 1.56 mole

$n_2$ = final moles of gas = ?

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

$\frac{V}{2V}=\frac{1.56mole}{n_2}$

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Now we have to calculate the mass of helium gas at doubled volume.

$\text{Mass of }He=\text{Moles of }He\times \text{Molar mass of }He$

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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.

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

<u>STEP 5:</u> 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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