
How many moles of $N{H_3}$are there in $250c{m^3}$ of a $30\% $ solution, the specific gravity of which is\[0.90\]?
A. \[3.97{\text{ }}moles\]
B. \[6.37{\text{ }}moles\]
C. \[3.70{\text{ }}moles\]
D. \[3.50{\text{ }}moles\]
Answer
483.6k+ views
Hint:We will take the help of the formula of Mass percentage. It is the amount of solute present in $100g$ 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$ = \dfrac{{Mass{\text{ of solute}}}}{{Mass{\text{ of solution}}}} \times 100$
Complete answer:
Given: Volume of water $ = 250ml$
Density of water$ = 0.999g/ml$
Since density $ = \dfrac{{Mass{\text{ of water}}}}{{Volume{\text{ of water}}}}$
$
6.999 = \dfrac{{Mass{\text{ of water}}}}{{250}} \\
\therefore Mass{\text{ of water}} = 0.999 \times 250 \\
$
$ = 249.75gm$
Since ammonia dissolved in ${H_2}O$ is \[30\% \]by weight
By using the formula, mass percentage$ = \dfrac{{Mass{\text{ of solute}}}}{{Mass{\text{ of solution}}}} \times 100$
Mass percentage $ = \dfrac{{Mass{\text{ of solute}}}}{{Mass{\text{ of solute}} + mass{\text{ of solvent}}}} \times 100$
$\therefore $$Mass{\text{ percentage}}$$ = \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\% $ and mass of water is $249.75gm,$by putting the values we will get-
$\dfrac{{30}}{{100}} = \dfrac{x}{{x + 249.75}}$
$
30x + 7492.5 = 100x \\
70x = 7492.5{\text{ }} \\
\therefore {\text{x = 107}}{\text{.03gm}}{\text{.}} \\
$
Thus mass of ammonia dissolved in \[250ml\] of water is \[107.03gm\]
To find the number of moles of ammonia we will use this formula
Number moles of $N{H_3} = \dfrac{{Mass{\text{ of ammonia}}}}{{Molar{\text{ mass}}}}$
\[
= \dfrac{{107.03g}}{{17g/mol}} \\
= 6.3moles \\
\]
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.
Formula used: Mass percentage$ = \dfrac{{Mass{\text{ of solute}}}}{{Mass{\text{ of solution}}}} \times 100$
Complete answer:
Given: Volume of water $ = 250ml$
Density of water$ = 0.999g/ml$
Since density $ = \dfrac{{Mass{\text{ of water}}}}{{Volume{\text{ of water}}}}$
$
6.999 = \dfrac{{Mass{\text{ of water}}}}{{250}} \\
\therefore Mass{\text{ of water}} = 0.999 \times 250 \\
$
$ = 249.75gm$
Since ammonia dissolved in ${H_2}O$ is \[30\% \]by weight
By using the formula, mass percentage$ = \dfrac{{Mass{\text{ of solute}}}}{{Mass{\text{ of solution}}}} \times 100$
Mass percentage $ = \dfrac{{Mass{\text{ of solute}}}}{{Mass{\text{ of solute}} + mass{\text{ of solvent}}}} \times 100$
$\therefore $$Mass{\text{ percentage}}$$ = \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\% $ and mass of water is $249.75gm,$by putting the values we will get-
$\dfrac{{30}}{{100}} = \dfrac{x}{{x + 249.75}}$
$
30x + 7492.5 = 100x \\
70x = 7492.5{\text{ }} \\
\therefore {\text{x = 107}}{\text{.03gm}}{\text{.}} \\
$
Thus mass of ammonia dissolved in \[250ml\] of water is \[107.03gm\]
To find the number of moles of ammonia we will use this formula
Number moles of $N{H_3} = \dfrac{{Mass{\text{ of ammonia}}}}{{Molar{\text{ mass}}}}$
\[
= \dfrac{{107.03g}}{{17g/mol}} \\
= 6.3moles \\
\]
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.
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