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Question: How many moles of \(N{H_3}\)are there in \(250c{m^3}\) of a \(30\% \) solution, the specific gravity...

How many moles of NH3N{H_3}are there in 250cm3250c{m^3} of a 30%30\% solution, the specific gravity of which is0.900.90?
A. 3.97 moles3.97{\text{ }}moles
B. 6.37 moles6.37{\text{ }}moles
C. 3.70 moles3.70{\text{ }}moles
D. 3.50 moles3.50{\text{ }}moles

Explanation

Solution

We will take the help of the formula of Mass percentage. It is the amount of solute present in 100g100g of solution. By this we can find the number of moles i.e. the number of moles is the ratio of given weight of solute to the molar mass of the solute.
Formula used: Mass percentage=Mass of soluteMass of solution×100 = \dfrac{{Mass{\text{ of solute}}}}{{Mass{\text{ of solution}}}} \times 100

Complete answer:
Given: Volume of water =250ml = 250ml
Density of water=0.999g/ml = 0.999g/ml
Since density =Mass of waterVolume of water = \dfrac{{Mass{\text{ of water}}}}{{Volume{\text{ of water}}}}
6.999=Mass of water250 Mass of water=0.999×250  6.999 = \dfrac{{Mass{\text{ of water}}}}{{250}} \\\ \therefore Mass{\text{ of water}} = 0.999 \times 250 \\\
=249.75gm= 249.75gm
Since ammonia dissolved in H2O{H_2}O is 30%30\% by weight
By using the formula, mass percentage=Mass of soluteMass of solution×100 = \dfrac{{Mass{\text{ of solute}}}}{{Mass{\text{ of solution}}}} \times 100
Mass percentage =Mass of soluteMass of solute+mass of solvent×100 = \dfrac{{Mass{\text{ of solute}}}}{{Mass{\text{ of solute}} + mass{\text{ of solvent}}}} \times 100
\therefore Mass percentageMass{\text{ percentage}} =Mass of ammoniaMass of ammonia+mass of water×100 = \dfrac{{Mass{\text{ of ammonia}}}}{{Mass{\text{ of ammonia}} + mass{\text{ of water}}}} \times 100
Now, let the mass of ammonia be mass percentage is 30%30\% and mass of water is 249.75gm,249.75gm,by putting the values we will get-
30100=xx+249.75\dfrac{{30}}{{100}} = \dfrac{x}{{x + 249.75}}
30x+7492.5=100x 70x=7492.5  x = 107.03gm.  30x + 7492.5 = 100x \\\ 70x = 7492.5{\text{ }} \\\ \therefore {\text{x = 107}}{\text{.03gm}}{\text{.}} \\\
Thus mass of ammonia dissolved in 250ml250ml of water is 107.03gm107.03gm
To find the number of moles of ammonia we will use this formula
Number moles of NH3=Mass of ammoniaMolar massN{H_3} = \dfrac{{Mass{\text{ of ammonia}}}}{{Molar{\text{ mass}}}}

=107.03g17g/mol =6.3moles  = \dfrac{{107.03g}}{{17g/mol}} \\\ = 6.3moles \\\

Hence the correct answer is option B.

Note: it is very important to know how the amount of substance is expressed when it is present in the form of a solution. The concentration of a solution or the amount of substance present in its given volume can be expressed in any of the following ways: mass percentage, mole fraction, molality and normality. Mass percentage is independent of temperature.