Question

In a laboratory experiment, a 10.0-mL sample of $\mathrm{NaCl}$solution is poured into an evaporating dish with a mass of $24.10\mathrm{g}$. The combined mass of the evaporating dish and $\mathrm{NaCl}$solution is $36.15\mathrm{g}$. After heating, the evaporating dish and dry $\mathrm{NaCl}$have a combined mass of $25.50\mathrm{g},(9.4)$

a. What is the mass percent$(\mathrm{m}/\mathrm{m})$of the $\mathrm{NaCl}$solution?

b. What is the molarity (M) of the $\mathrm{NaCl}$solution?

c. If water is added to $10.0\mathrm{mL}$of the initial $\mathrm{NaCl}$solution to give a final volume of $60.0\mathrm{mL}$, what is the molarity of the diluted $\mathrm{NaCl}$ solution?

Short answer

Part a) The mass percentage is 11.6 percent NaCl solution.

Part b) The molarity of the NaCl solution is 2.39 M.

Part C) The molarity of the diluted The solution for NaCl is 0.398 M.

Step-by-step solution

Step - 1   Introduction (Part a)

The mass of the $\mathrm{NaCl}$solution is obtained by subtracting the mass of the evaporating dish from the combined mass of the evaporating dish and the $\mathrm{NaCl}$solution.

Step -2  Given information (part-a)

(a)

• The sample $\mathrm{NaCl}$solution has a volume of $10.0\mathrm{mL}$.
• The evaporating dish has a mass of $24.10\mathrm{g}$.
• The evaporating dish and the $\mathrm{NaCl}$solution have a total mass of $36.15\mathrm{g}$.
• The total mass of the evaporating dish and dry $\mathrm{NaCl}$ is 25.50 g.

Step - 3  Explanation (part-a)

• According to the statistics, the mass of the evaporating dish is and the total mass of the evaporating dish and the solution is 36.15 g.
• Subtracting the mass of the evaporating dish from the combined mass of the evaporating dish and the $\mathrm{NaCl}$solution yields the mass of the $\mathrm{NaCl}$solution.

solution mass of NaCl $\left(g\right)=36.15g-24.10g$

$=12.05\mathrm{g}$

• Subtracting the mass of the evaporating dish from the combined mass of the evaporating dish and dry $\mathrm{NaCl}$yields the mass of dry $\mathrm{NaCl}$.

dry mass NaCl mass (g) $=25.50\mathrm{g}-24.10\mathrm{g}$

$=1.40\mathrm{g}$

• For exactly 100 g of solution, the mass percent (m/m) indicates the concentration of the solution in terms of the mass of the solute in grams.

$\text{Mass percent}(\mathrm{m}/\mathrm{m})$= $\frac{\text{mass of solute}\left(\mathrm{g}\right)}{\text{mass of solution}\left(\mathrm{g}\right)}\times 100\%$

• We may calculate the mass percent of the $\mathrm{NaCl}$solution by putting the mass of the solute and the mass of the solution into the mass percent calculation.
• As a result, the mass percent of the $\mathrm{NaCl}$solution is$(\mathrm{m}/\mathrm{m})$.

$\text{Mass percent}(\mathrm{m}/\mathrm{m})=\frac{\text{mass of solute}\left(\mathrm{g}\right)}{\text{mass of solution}\left(\mathrm{g}\right)}\times 100\%$

$=\frac{1.40\mathrm{gNaCl}}{12.05\mathrm{g}\text{solution}}\times 100\%$

$=11.6\%(\mathrm{m}/\mathrm{m})\mathrm{NaCl}\text{solution}$

• As a result, the mass percentage is 11.6 percent NaCl solution .

Step - 4  given information (part-b)

(b)

• Molarity (M) is a concentration unit measured in moles of solute per litre of solution.

$\text{Molarity}\left(\mathrm{M}\right)=\frac{\text{moles of solute}}{\text{liters of solution}}$

• We must first determine the moles of$\mathrm{NaCl}$ and the volume of the solution in liters in order to establish the molarity of the solution.

Step - 5  Explanation (part- b)

The volume of the sample $\mathrm{NaCl}$solution is 10.0 L. To convert $\mathrm{mL}\text{to}\mathrm{L}$ , we first utilize the conversion factor for following equality.

$1,000\mathrm{mL}=1\mathrm{L}$

$10.0\mathrm{mL}\mathrm{NaCl}\text{solution}\times \frac{1\mathrm{L}}{1,000\mathrm{mL}}=1.00\times {10}^{-2}\mathrm{L}\mathrm{NaCl}\text{solution}$

• $\mathrm{NaCl}$moles have now been calculated. The mass of dry$\mathrm{NaCl}$in the sample solution is 1.40 g.
• $\mathrm{NaCl}$has a molar mass of 58.5 g /mol.
• As a result, the following equivalence exists.

$1\mathrm{molNaCl}=58.5\mathrm{gNaCl}$

• We calculate the moles of $\mathrm{NaCl}\text{.}$using this equality as a conversion factor.
• As a result, we've
  $\text{Moles of}\mathrm{NaCl}=1.40-\mathrm{gNaCl}\times \frac{1\mathrm{molNaCl}}{58.5\mathrm{gNaCl}}$

$=0.0239\mathrm{molNaCl}$

• The molarity of the NaCl solution is

$\text{Molarity}\left(M\right)=\frac{\text{moles of solute}}{\text{liters of solution}}$

$=\frac{0.0239\mathrm{molNaCl}}{1.00\times {10}^{-2}\mathrm{LNaCl}\text{solution}}$

$=2.39\mathrm{M}$

As a result, the molarity of the NaCl solution is 2.39 M .

Step - 6  Given information (part-c)

(c)

• The concentration of the diluted solution will be calculated using the dilution equation. The following is the dilution formula:

${\mathrm{C}}_1{V}_1={\mathrm{C}}_2{V}_2$

• The concentration and volume of the concentrated solution are represented by C1and V1 .
• C 2 The concentration and volume of the diluted solution are shown by text and V 2 text.

Step - 7  Explanation (part-c)

• The concentration of 60.0 ML was calculated using the aforementioned dilution equation.
• NaCl solution created from 10.0 mL of initial sample NaCl solution can be computed.
• We know that the initial NaCl solution has a molarity of 2.39 M. Here,

$C_1=2.39\mathrm{M}$

$C_2=?M$
$V_1=10.0\mathrm{mL}$

$V_2=60.0\mathrm{mL}$

To determine${\mathrm{C}}_2$, the dilution expression is rearranged.

${\mathrm{C}}_1{V}_1={\mathrm{C}}_2{V}_2$

$\frac{{\mathrm{C}}_1{V}_1}{V_2}=\frac{{\mathrm{C}}_2{V}_2}{V_2}$

Therefore,

${\mathrm{C}}_2={\mathrm{C}}_1\times \frac{V_1}{V_2}$

• We now have by substituting the known quantities into the previous equation and solving for

${\mathrm{C}}_2=2.39\mathrm{M}\times \frac{10.0\mathrm{mL}}{60.0\mathrm{mL}}$

$=0.398\mathrm{M}$

• As a result, the molarity of the diluted The solution for NaCl is 0.398 M .