Question

Calculate the amount of the water ‘in $mL$’ which must be added to a given solution of concentration of $40mg$ silver nitrate per $1ml$, to yield a solution of concentration of $16mg$ silver nitrate per $1ml$ ?

Answer 1

Hint: As the amount of solute (silver nitrate) present is the same in both the initial and final solutions, we can equate this on both sides. The amount of water added will be attained by equating the strength of the initial solution and its volume ($1ml$) with the strength and volume added of the final solution.

Complete step by step answer:
The process involved here is dilution. To find the amount of water added, we can make use of the molarity-volume equation, which states that the product of molarity and volume is a constant:
${M_1}{V_1} = {M_2}{V_2}$
Where ${M_1},{M_2}$ are the initial and final molarities and ${V_1},{V_2}$ are the initial and final volumes of water added respectively.
Initially, let us consider the case where the volume of the solution is $1ml$. We know, molarity:
$M = \dfrac{{{n_A}}}{V}$ where ${n_A}$ is the number of moles of the solute and $V$ is the volume of the solution.
And ${n_A} = \dfrac{W}{m}$ where $W$ is the given mass and $m$ is the molar mass of the solute.
Here given mass is $40mg = 40 \times {10^{ - 3}}g$, while molar mass of silver nitrate ($AgN{O_3}$) is $108 + 14 + (16 \times 3) = 170$. Therefore, number of moles of $AgN{O_3}$:
${n_A} = \dfrac{{40 \times {{10}^{ - 3}}}}{{170}}$
Since we have taken volume as $1mL = {10^{ - 3}}L$, initial molarity is:
${M_1} = \dfrac{{\dfrac{{40 \times {{10}^{ - 3}}}}{{170}}}}{{{{10}^{ - 3}}}} = \dfrac{{40}}{{170}}mol/L$
Final solution contains $16mg$ per $mL$. Thus, number of moles is:
${n_A} = \dfrac{{16 \times {{10}^{ - 3}}}}{{170}}$ And hence, molarity of the final solution:
${M_2} = \dfrac{{\dfrac{{16 \times {{10}^{ - 3}}}}{{170}}}}{{{{10}^{ - 3}}}} = \dfrac{{16}}{{170}}mol/L$
Let the water to be added be $'V'mL$. Thus, since the initial volume as we have taken earlier is $1ml$, final volume is $(V + 1)mL$. And we have got that ${M_1} = \dfrac{{40}}{{170}}$ and ${M_2} = \dfrac{{16}}{{170}}mol/L$. Putting these values into our molarity-volume equation, we get:
$\left( {\dfrac{{40}}{{170}}mol/L} \right) \times 1mL = \left( {\dfrac{{16}}{{170}}mol/L} \right)(V + 1)$
Cancelling off the denominators on both sides and rearranging, we get:
$V + 1 = \dfrac{{40}}{{16}} = 2.5$
Hence, $V = 2.5 - 1 = 1.5mL$
Therefore, the volume of water to be added is $1.5mL$.

Additional information: Silver nitrate is an inorganic compound and is used as a substituent for many other silver chemical substances. It is very less sensitive to light than the halides. It is also used as an antiseptic to its antibacterial activity. Used in the treatment of burns and wounds on the skin.

Note:The concentration term unit which we used here gives us the weight of the solute (silver nitrate) per unit volume of the solution. Another commonly used concentration unit is molarity, which gives us the number of moles per unit volume. The unit molality on the other hand, gives us the number of moles of solute per unit mass of the solvent. The number of moles of the solute is found by dividing the mass of silver nitrate given with its molecular mass.