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Question: An ammonia solution has a density of \(0.910gc{m^{ - 3}}\) and is \(25\% \) ammonia by mass. What is...

An ammonia solution has a density of 0.910gcm−30.910gc{m^{ - 3}} and is 25%25\% ammonia by mass. What is the molarity of the solution?
A.12.1M12.1M
B.13.4M13.4M
C.14.5M14.5M
D.15.5M15.5M

Explanation

Solution

There are many ways of expressing the concentration of the solution. Molarity is one of them. It gives the relation between the mass of solute and volume of solution. We also know that the density of any compound is related to its mass.

Complete step by step answer:
We know that we can express the concentration of solution in many ways. Normality, molarity, molality are some of them.
We know that a solution is made of solute and solvent. The component present in major proportion is called solvent while the minor component is called as solute.
Molarity can be defined as the number of moles of solute dissolved per litre of the solution. It is denoted by MM . The formula can be given as-
Molarity =number of moles of soluteVolume of solution in litres = \dfrac{{number\,of\,moles\,of\,solute}}{{Volume\,of\,solution\,in\,litres}}
Now in the given problem we have to find the molarity of the solution. So the volume of solution is 1litres1litres . We need to calculate the number of moles of ammonia.
Number of moles can be calculated as follows-
Number of moles of ammonia =given mass of ammoniamolar mass of ammonia = \dfrac{{given\,mass\,of\,ammonia}}{{molar\,mass\,of\,ammonia}}
Now the density and mass percentage of ammonia is given.
The density of ammonia is 0.910gcm−30.910gc{m^{ - 3}} per litre .
We know 1litre=1000cm−31litre = 1000c{m^{ - 3}}
So, 1000cm−3×0.910gcm−3=0.910g1000c{m^{ - 3}} \times 0.910gc{m^{ - 3}} = 0.910g
So the solution will have 0.910g0.910g of ammonia per litre.
But it is given that there is 25%25\% ammonia by mass in the solution. So the mass of ammonia in solution of density 0.910gcm−30.910gc{m^{ - 3}} will be-
Mass of ammonia =25100×0.910=227.5g = \dfrac{{25}}{{100}} \times 0.910 = 227.5g of ammonia.
Now the number of moles of ammonia in 227.5g227.5g of ammonia will be-
Number of moles =227.517=13.4 = \dfrac{{227.5}}{{17}} = 13.4 (since molar mass of ammonia is 17gmol−117gmo{l^{ - 1}} .
So molarity is-
Molarity =13.41=13.4M = \dfrac{{13.4}}{1} = 13.4M
The molarity of the given solution is 13.4M13.4M .
The correct option is B.

Note: Molarity is a function of temperature. Because the volume of solution depends on temperature.
-If the solvent is 1kg1kg then we use the term molality. It is defined as the number of moles of solute per kgkg of solvent. Molality is denoted by mm .
-Molality does not depend on temperature because mass does not change with temperature.