Question

The following five beakers, each containing a solution of sodium chloride (NaCl, also known as table salt), were found on a lab shelf:
Beaker Contents
1 200 mL. of 1.50 M NaCl solution
2 100 mL. of 3.00 M NaCl solution
3 150 mL. of solution containing 22.5 g of NaCl
4 100 mL. of solution containing 22.5 g of NaCl
5 300 mL. of solution containing 0.450 mol NaCl

Arrange the solutions in order of decreasing concentration.
Rank from most concentrated to least concentrated. To rank items as equivalent, overlap them.

Part B:
A student placed 10.5 of glucose (C6H12O6) in a volumetric flask, added enough water to dissolve the glucose by swirling, then carefully added additional water until the 100.-mL mark on the neck of the flask was reached. The flask was then shaken until the solution was uniform. A 20.0-mL sample of this glucose solution was diluted to 0.500L . How many grams of glucose are in 100.mL of the final solution?
Express your answer numerically in grams.
S

Answer 1

Concepts and reason

The problem is based on the concept of molarity of a solution.

The concentration of solution is referred to as molarity of solution. It is basically amount of solute in given amount of solvent. If in a solution, amount of solute is more than that of solvent then solution is said to be concentrated. For more amount of solvent than solute, solution is said to be diluted.

Dilution is the process of adding more solvent to a solution in order to reduce the concentration of solute. During dilution, the concentration of solute decreases, whereas the volume of the solution increases.

Fundamentals

Molarity is a unit which is used to express the relative concentration of solutes and solvent.

It is defined as number of moles of solute per liters of solution. It is denoted by letter M.

It is mathematically represented as follows:

${\rm{M}} = \frac{{{\rm{Moles of solute}}}}{{{\rm{Volume of solution in L}}}}$

Therefore, Unit of molarity is ${\rm{mol }}{{\rm{L}}^{ - 1}}$ or M.

A dilution equation is used to calculate the concentration of solute or the volume of the solution. It is given by the following expression:

${{\rm{M}}_1}{{\rm{V}}_1} = {{\rm{M}}_2}{{\rm{V}}_2}$

Here, ${{\rm{M}}_1}$ is initial molarity, ${{\rm{M}}_2}$ is final molarity, ${{\rm{V}}_1}$ is initial volume and ${{\rm{V}}_2}$is final volume.

Part A

Solution 1: In 200 mL of 1.50 M NaCl solution, the concentration of solution is equal to 1.50 M.

Solution 2: In 100 mL of 3.00 M NaCl solution, the concentration of solution is equal to

3.00 M.

Solution 3: 150 mL of solution containing 22.5 g of NaCl.

Calculate number of moles of NaCl in solution as follows:

$n = \frac{m}{{{\rm{M}}{\rm{.M}}}}$ …… (1)

Here, m is mass of NaCl and M.M is molar mass of NaCl.

Substitute 22.5 g for m and $58.5{\rm{ g mo}}{{\rm{l}}^{ - 1}}$ for M.M thus,

$\begin{array}{c}\\n = \frac{{22.5{\rm{ g}}}}{{58.5{\rm{ g mo}}{{\rm{l}}^{ - 1}}}}\\\\ = 0.3846{\rm{ mol}}\\\end{array}$

Calculate molarity of solution as follows:

${\rm{M}} = \frac{{{\rm{Moles of solute}}}}{{{\rm{Volume of solution in L}}}}$

Substitute $0.3846{\rm{ mol}}$ for moles of solute and 150 mL for volume of solution thus,

$\begin{array}{c}\\{\rm{M}} = \frac{{0.3846{\rm{ mol}}}}{{150{\rm{ mL}}}}\left( {\frac{{1{\rm{ mL}}}}{{{{10}^{ - 3}}{\rm{ L}}}}} \right)\\\\ = 2.564{\rm{ mol }}{{\rm{L}}^{ - 1}}\\\\ = 2.564{\rm{ M}}\\\end{array}$

Solution 4: 100mL of solution containing 22.5g of NaCl.

Substitute 22.5 g for m and $58.5{\rm{ g mo}}{{\rm{l}}^{ - 1}}$ for M.M in equation (1) thus,

$\begin{array}{c}\\n = \frac{{22.5{\rm{ g}}}}{{58.5{\rm{ g mo}}{{\rm{l}}^{ - 1}}}}\\\\ = 0.3846{\rm{ mol}}\\\end{array}$

Calculate molarity of solution as follows:

${\rm{M}} = \frac{{{\rm{Moles of solute}}}}{{{\rm{Volume of solution in L}}}}$

Substitute $0.3846{\rm{ mol}}$ for moles of solute and 100 mL for volume of solution thus,

$\begin{array}{c}\\{\rm{M}} = \frac{{0.3846{\rm{ mol}}}}{{100{\rm{ mL}}}}\left( {\frac{{1{\rm{ mL}}}}{{{{10}^{ - 3}}{\rm{ L}}}}} \right)\\\\ = 3.846{\rm{ mol }}{{\rm{L}}^{ - 1}}\\\\ = 3.846{\rm{ M}}\\\end{array}$

Solution 5: 300 mL of solution containing 0.450 mol of NaCl.

Calculate molarity of solution as follows:

${\rm{M}} = \frac{{{\rm{Moles of solute}}}}{{{\rm{Volume of solution in L}}}}$

Substitute $0.450{\rm{ mol}}$ for moles of solute and 300 mL for volume of solution thus,

$\begin{array}{c}\\{\rm{M}} = \frac{{0.450{\rm{ mol}}}}{{300{\rm{ mL}}}}\left( {\frac{{1{\rm{ mL}}}}{{{{10}^{ - 3}}{\rm{ L}}}}} \right)\\\\ = 1.5{\rm{ mol }}{{\rm{L}}^{ - 1}}\\\\ = 1.5{\rm{ M}}\\\end{array}$

The decreasing order of the concentrations of NaCl is as follows:

$3.846{\rm{M}} > {\rm{3}}{\rm{.00M}} > 2.564{\rm{M}} > 1.5{\rm{M}} = {\rm{1}}{\rm{.5M}}$

Part B

In 0.1 L of stock solution, mass of glucose present is equals to 10.5 g.

Fist calculate number of moles of glucose as follows:

$n = \frac{m}{{{\rm{M}}{\rm{.M}}}}$

Here, m is mass of glucose and M.M is molar mass of glucose.

Substitute 10.5 g for m and $180{\rm{ g mo}}{{\rm{l}}^{ - 1}}$ for M.M thus,

$\begin{array}{c}\\n = \frac{{10.5{\rm{ g}}}}{{180{\rm{ g mo}}{{\rm{l}}^{ - 1}}}}\\\\ = 0.0583{\rm{ mol}}\\\end{array}$

Now, calculate molarity, M of the solution as follows:

$M = \frac{n}{V}$

Substitute $0.0583{\rm{ mol}}$ for n and 100 mL for V thus,

$\begin{array}{c}\\M = \frac{{0.0583{\rm{ mol}}}}{{{\rm{100 mL}}}}\left( {\frac{{1{\rm{ mL}}}}{{{{10}^{ - 3}}{\rm{ L}}}}} \right)\\\\ = 0.583{\rm{ mol }}{{\rm{L}}^{ - 1}}\\\\ = 0.583{\rm{ M}}\\\end{array}$

Now, 20 mL of solution is taken therefore, initial volume, ${V_1}$ is 20 mL and initial molarity, ${M_1}$ is equals to 0.583 M. Calculate the molarity, ${M_2}$ of the diluted solution for volume, ${V_2}$ equals to 0.5 L using dilution equation as follows:

According to dilution Law,

${M_1}{V_1} = {M_2}{V_2}$

Or,

${M_2} = \frac{{{M_1}{V_1}}}{{{V_2}}}$

Substitute 0.583 M for ${{\rm{M}}_1}$, 20 mL for ${{\rm{V}}_1}$, 0.5 L for ${{\rm{V}}_2}$ thus,

$\begin{array}{c}\\{{\rm{M}}_2} = \frac{{{{\rm{M}}_1}{{\rm{V}}_1}}}{{{{\rm{V}}_2}}}\\\\ = \frac{{\left( {0.583{\rm{ M}}} \right)\left( {20{\rm{ mL}}} \right)}}{{\left( {0.5{\rm{ L}}} \right)}}\left( {\frac{{1{\rm{ L}}}}{{{{10}^3}{\rm{ mL}}}}} \right)\\\\ = 0.02332{\rm{M}}\\\end{array}$

Calculate number of moles of glucose from molarity as follows:

$n = M \times V$

Substitute ${\rm{0}}{\rm{.02332 M}}$ for M and 100 mL for V thus,

$\begin{array}{c}\\n = \left( {{\rm{0}}{\rm{.02332 M}}} \right)\left( {\frac{{{\rm{mol }}{{\rm{L}}^{ - 1}}}}{{1{\rm{ M}}}}} \right)\left( {100{\rm{ mL}}} \right)\left( {\frac{{{{10}^{ - 3}}{\rm{ L}}}}{{1{\rm{ mL}}}}} \right)\\\\ = 0.002332{\rm{ mol}}\\\end{array}$

Calculate mass of glucose as follows:

$m = n \times {\rm{M}}{\rm{.M}}$

Substitute $0.002332{\rm{ mol}}$for n and $180{\rm{ g mo}}{{\rm{l}}^{ - 1}}$ for M.M thus,

$\begin{array}{c}\\m = \left( {0.002332{\rm{ mol}}} \right)\left( {180{\rm{ g mo}}{{\rm{l}}^{ - 1}}} \right)\\\\ = {\rm{0}}{\rm{.419 g}}\\\end{array}$

Ans: Part A

The order of solutions in decreasing concentration is $4 > 2 > 3 > 1 = 5$

Part B

Mass of glucose in 100 mL is 0.419g.