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olya-2409 [2.1K]
3 years ago
7

The density of an aqueous solution of nitric acid is 1.64 g/mL and the concentration is 1.85 M. What is the concentration of thi

s solution in percent by mass?
Chemistry
1 answer:
galina1969 [7]3 years ago
6 0

Answer:

Mass % of the solution = 7.1067 %

Explanation:

Given :

Molarity of nitric acid solution = 1.85 M

Density of the solution = 1.64 g/mL

<u>Molarity of a solution is defined as the number of moles of solute present in 1 liter of the solution.</u>

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

Lets, consider the volume of the solution = 1 L

Thus,

Moles of nitric acid present in the solution:

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

Moles of Nitric acid=Molarity \times {Volume\ of\ the\ solution}

So,

Moles of Nitric acid  = 1.85 moles

Molar mass of nitric acid = 63 g/mol

The mass of Nitric acid can be find out by using mole formula as:

moles=\frac{Mass\ taken}{Molar\ mass}

Thus,  

Mass\ of\ Nitric\ acid=Moles \times Molar mass}

Mass\ of\ Nitric\ acid=1.85 g \times 63 g/mol}

<u>Mass of Nitric acid = 116.55 g</u>

Also,

Density=\frac{Mass}{Volume}

Given : Density = 1.64 g/mL

Also, 1 L = 10³ mL

Volume of the solution is 1000 mL

So, mass of the solution:

Mass\ of\ the\ solution=Density \times {Volume\ of\ the\ solution}

Mass\ of\ the\ solution=1.64 g/mL \times {1000 mL}

<u>Mass of the solution  = 1640 g</u>

Mass % is defined as the mass of solute in 100 g of the solution. The formula for the calculation of mass % is shown below:

Mass \% =\frac{Mass\ of\ the\ solute}{Mass\ of\ the\ solution} \times {100}

So,

Mass \%=\frac{116.55}{1640} \times {100}

<u>Mass % = 7.1067 %</u>

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15.89 percent carbon, 21.18 percent oxygen, and 62.93 percent osmium. what is the empirical formula
otez555 [7]

Answer:

OsCO or COOs

Explanation:

Data given

Carbon = 15.89 %

Oxygen = 21.18 %

Osmium = 62.93%

Empirical formula = ?

Solution:

First find the masses of each component

Consider total compound is 100g

As we now

mass of element = % of component

So,

15.89 g of C     = 15.89 % Carbon

21.18 g of O      =   21.18 % Oxygen

62.93 g of Os  =   62.93% Osmium

Now convert the masses to moles

For Carbon

Molar mass of C = 12 g/mol

                  no. of mole = mass in g / molar mass

Put value in above formula

                  no. of mole =  15.89 g/ 12 g/mol

                  no. of mole =  1.3242

mole of C = 1.3242

For Oxygen

Molar mass of O = 16 g/mol

                  no. of mole = mass in g / molar mass

Put value in above formula

                  no. of mole =  21.18 g/ 16 g/mol

                  no. of mole =  

mole of O = 1.3238

For Os

Molar mass of Os = 190 g/mol

                  no. of mole = mass in g / molar mass

Put value in above formula

                  no. of mole =  62.93 g/ 190 g/mol

                  no. of mole =  

mole of Os = 1.3312

Now we have values in moles as below

C = 1.3242

O = 1.3238

Os = 1.3312

Divide the all values on the smalest values to get whole number ratio

C = 1.3242 /1.3238 = 1.0003

O = 1.3238 /1.3238 = 1

Os = 1.3312 /1.3238 = 1.0056

So all have round value 1 mole

C = 1

O = 1

Os = 1

So the empirical formula will be (OsCO) i.e. all 3 atoms in simplest small ratio

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