Method of expressing concentration of solution Questions in English

(A) The density of pure water is approximately $1 \ g/mL$.
For $1 \ L$ $(1000 \ mL)$ of water,the mass is $1000 \ g$.
The molar mass of water $(H_2O)$ is $18 \ g/mol$.
Number of moles $(n)$ = $\frac{\text{mass}}{\text{molar mass}} = \frac{1000 \ g}{18 \ g/mol} \approx 55.56 \ mol$.
Molarity $(M)$ = $\frac{\text{moles of solute}}{\text{volume of solution in liters}} = \frac{55.56 \ mol}{1 \ L} = 55.56 \ M \approx 55.6 \ M$.

(C) Molarity $(M)$ is defined as the number of moles of solute per liter of solution.
First,calculate the number of moles of ethanol $(C_2H_5OH)$:
$\text{Moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{23 \ g}{46 \ g \ mol^{-1}} = 0.5 \ mol$.
Next,convert the volume of the solution from $mL$ to $L$:
$500 \ mL = 0.5 \ L$.
Finally,calculate the molarity:
$M = \frac{\text{moles of solute}}{\text{volume of solution in } L} = \frac{0.5 \ mol}{0.5 \ L} = 1 \ M$.

(D) Moles of water $= \frac{x}{18}$.
Moles of ethanol $= \frac{69}{46} = 1.5$.
Mole fraction of ethanol $(X_{ethanol}) = \frac{n_{ethanol}}{n_{water} + n_{ethanol}}$.
Given $X_{ethanol} = 0.6$,so $0.6 = \frac{1.5}{\frac{x}{18} + 1.5}$.
$\frac{x}{18} + 1.5 = \frac{1.5}{0.6} = 2.5$.
$\frac{x}{18} = 2.5 - 1.5 = 1.0$.
$x = 18 \, g$.

(B) solution with molality $x \ m$ means $x \ mol$ of solute is present in $1000 \ g$ of solvent (benzene).
Moles of solute $= x$
Moles of benzene $= \frac{1000}{78} \approx 12.82 \ mol$ (Molar mass of benzene $= 78 \ g \ mol^{-1}$).
Mole fraction of solute $(X_{solute}) = 0.2$.
Formula for mole fraction: $X_{solute} = \frac{n_{solute}}{n_{solute} + n_{solvent}}$.
$0.2 = \frac{x}{x + 12.82}$.
$0.2(x + 12.82) = x$.
$0.2x + 2.564 = x$.
$0.8x = 2.564$.
$x = \frac{2.564}{0.8} = 3.205 \approx 3.2$.

(D) The density of $HCl$ is $1.17 \ g/cm^3$,which means $1 \ cm^3$ $(1 \ mL)$ of the solution contains $1.17 \ g$ of $HCl$.
The molar mass of $HCl$ is $36.5 \ g/mol$.
Number of moles in $1 \ L$ $(1000 \ mL)$ of solution = $\frac{1.17 \ g}{36.5 \ g/mol} \times 1000 \ mL = 32.05 \ mol$.
Therefore,the molarity is $32.05 \ M$.

(A) Given,mole fraction of water $X_{H_2O} = 0.85$.
Therefore,mole fraction of $H_2SO_4$ is $X_{H_2SO_4} = 1 - 0.85 = 0.15$.
Molality $(m)$ is defined as the number of moles of solute per $1000 \ g$ of solvent.
Moles of water $(n_{H_2O})$ in $1000 \ g$ of water = $\frac{1000 \ g}{18 \ g/mol} = 55.55 \ mol$.
Using the relation $\frac{n_{H_2SO_4}}{n_{H_2O}} = \frac{X_{H_2SO_4}}{X_{H_2O}}$:
$n_{H_2SO_4} = \frac{0.15}{0.85} \times 55.55 = 9.80 \ mol$.
Since this amount of solute is present in $1000 \ g$ of solvent,the molality is $9.80 \ m$.

(D) Using the dilution formula $N_1V_1 = N_2V_2$:
Given $N_1 = 10 \,N$,$N_2 = 1 \,N$,and $V_2 = 1000 \,cc$.
Substituting the values: $10 \,N \times V_1 = 1 \,N \times 1000 \,cc$.
$V_1 = \frac{1000}{10} \,cc = 100 \,cc$.
Therefore,$100 \,cc$ of concentrated $HCl$ is required to prepare $1000 \,cc$ of $1 \,N \,HCl$.

(B) The molar mass of water $(H_2O)$ is $18 \ g/mol$.
Number of moles of water = $\frac{1000 \ g}{18 \ g/mol} \approx 55.55 \ mol$.
Number of moles of urea = $1 \ mol$.
Total number of moles = $55.55 + 1 = 56.55 \ mol$.
Mole fraction of water $(x_{water})$ = $\frac{\text{moles of water}}{\text{total moles}} = \frac{55.55}{56.55} \approx 0.982$.

(B) The mole fraction $(x)$ of a component in a solution is defined as the ratio of the number of moles of that component to the total number of moles of all components present in the solution.
Since the number of moles of any component cannot be negative and cannot exceed the total number of moles,the value of the mole fraction must lie between $0$ and $1$ inclusive,i.e.,$0 \leq x \leq 1$.
Therefore,the statement $-2 \leq x \leq 2$ is incorrect.

(D) The molarity of the mixture is calculated using the formula: $M_{mixture} = \frac{M_1V_1 + M_2V_2}{V_1 + V_2}$
Substituting the given values:
$M_{mixture} = \frac{(1.5 \times 480) + (1.2 \times 520)}{480 + 520}$
$M_{mixture} = \frac{720 + 624}{1000}$
$M_{mixture} = \frac{1344}{1000} = 1.344 \ M$

(A) The molar mass of ethylene glycol $[C_2H_4(OH)_2]$ is $(2 \times 12) + (6 \times 1) + (2 \times 16) = 62 \ g/mol$.
Molality $(m)$ is defined as the number of moles of solute per kilogram of solvent.
Moles of ethylene glycol $= \frac{222.6 \ g}{62 \ g/mol} = 3.59 \ mol$.
Mass of solvent (water) $= 200 \ g = 0.2 \ kg$.
Molality $(m) = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{3.59 \ mol}{0.2 \ kg} = 17.95 \ m$.

(A) Concentration terms that involve volume are temperature-dependent because volume changes with temperature.
$Molarity (M)$ is defined as the number of moles of solute per liter of solution.
Since volume of the solution changes with temperature,$Molarity$ also changes with temperature.
Other terms like $Molality$,$Mole fraction$,and $Mass percentage$ involve mass,which is independent of temperature.

(B) Given: Molarity $(M) = 2.05 \ mol \ L^{-1}$,Density $(d) = 1.02 \ g \ mL^{-1}$.
Molar mass of acetic acid $(CH_3COOH) = 12 + 3 + 12 + 16 + 16 + 1 = 60 \ g \ mol^{-1}$.
Mass of solute in $1 \ L$ solution = $2.05 \ mol \times 60 \ g \ mol^{-1} = 123 \ g$.
Mass of $1 \ L$ solution = $Volume \times Density = 1000 \ mL \times 1.02 \ g \ mL^{-1} = 1020 \ g$.
Mass of solvent (water) = $Mass \ of \ solution - Mass \ of \ solute = 1020 \ g - 123 \ g = 897 \ g = 0.897 \ kg$.
Molality $(m) = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{2.05 \ mol}{0.897 \ kg} \approx 2.28 \ mol \ kg^{-1}$.

(B) Volume of solute (isopropyl alcohol) = $125 \, cm^3$.
Total volume of solution = $175 \, cm^3$.
Volume fraction = $\frac{\text{Volume of solute}}{\text{Total volume of solution}} = \frac{125}{175} = 0.714$.
Volume percentage = $\text{Volume fraction} \times 100 = 0.714 \times 100 = 71.4\%$.
Thus,the volume fraction is $0.714$ and the volume percentage is $71.4\%$.

(C) The dilution formula is $N_1V_1 = N_2V_2$.
Here,$N_1 = 0.5 \ N$ (seminormal),$V_1 = 200 \ cc$.
$N_2 = 0.1 \ N$ (decinormal),$V_2 = ?$.
Substituting the values: $0.5 \times 200 = 0.1 \times V_2$.
$V_2 = \frac{0.5 \times 200}{0.1} = 1000 \ cc$.
The volume of water to be added is $V_2 - V_1 = 1000 - 200 = 800 \ cc$.

(C) Using the dilution formula: $N_1V_1 = N_2V_2$
Here,$N_1 = 36 \ N$,$V_1 = 50 \ mL$,$V_2 = 50 \ mL + 50 \ mL = 100 \ mL$.
$36 \times 50 = N_2 \times 100$
$N_2 = \frac{1800}{100} = 18 \ N$.
For $H_2SO_4$,the n-factor $(n)$ is $2$.
Since $M = \frac{N}{n}$,we have $M = \frac{18}{2} = 9 \ M$.

(D) The number of moles of $HCl$ $(n)$ is calculated as: $n = \frac{w}{m} = \frac{3.65 \ g}{36.5 \ g/mol} = 0.1 \ mol$.
The number of moles of water $(N)$ is calculated as: $N = \frac{W}{M} = \frac{16.2 \ g}{18 \ g/mol} = 0.9 \ mol$.
The mole fraction of $HCl$ $(X_{HCl})$ is given by: $X_{HCl} = \frac{n}{n + N} = \frac{0.1}{0.1 + 0.9} = \frac{0.1}{1.0} = 0.1$.

(C) First,calculate the molarity $(M_1)$ of the stock solution using the formula: $M = \frac{\% \, (w/w) \times d \times 10}{Molar \, Mass}$.
$M_1 = \frac{29.2 \times 1.25 \times 10}{36.5} = \frac{365}{36.5} = 10 \, M$.
Now,use the dilution equation: $M_1V_1 = M_2V_2$.
$10 \, M \times V_1 = 0.4 \, M \times 200 \, mL$.
$V_1 = \frac{0.4 \times 200}{10} = 8 \, mL$.

(B) Given: Mass of urea $= 0.0100 \ g$.
Molecular mass of urea $[(NH_2)_2CO] = 60 \ g/mol$.
Number of moles of urea $= \frac{0.0100 \ g}{60 \ g/mol} = 1.667 \times 10^{-4} \ mol$.
Volume of water $= 0.3000 \ m^3 = 300 \ L = 300,000 \ mL$.
Since density of water is $1 \ g/mL$,mass of water $= 300,000 \ g = 300 \ kg$.
Molality $(m) = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{1.667 \times 10^{-4} \ mol}{300 \ kg} = 5.55 \times 10^{-7} \ m$.
Note: Based on the provided options,the calculation assumes the volume was $0.3000 \ L$ instead of $0.3000 \ m^3$. If volume $= 0.3000 \ L$,then mass $= 0.3 \ kg$,and $m = \frac{1.667 \times 10^{-4}}{0.3} = 5.55 \times 10^{-4} \ m$.

(D) The concentration in $ppm$ is calculated as:
$ppm = \frac{\text{mass of solute}}{\text{mass of solution}} \times 10^6$
$ppm = \frac{10 \ g}{1000 \ g} \times 10^6$
$ppm = 0.01 \times 10^6 = 10,000 \ ppm$

(A) $10\% (w/v)$ solution means $10 \, \text{g}$ of glucose is present in $100 \, \text{mL}$ of solution.
The molar mass of glucose $(C_6H_{12}O_6)$ is $180 \, \text{g/mol}$.
Mass of $2 \, \text{moles}$ of glucose $= 2 \, \text{mol} \times 180 \, \text{g/mol} = 360 \, \text{g}$.
Since $10 \, \text{g}$ of glucose is in $100 \, \text{mL}$ of solution,then $360 \, \text{g}$ of glucose will be in:
$\text{Volume} = \frac{100 \, \text{mL}}{10 \, \text{g}} \times 360 \, \text{g} = 3600 \, \text{mL}$.
Converting to liters: $3600 \, \text{mL} = 3.6 \, \text{L}$.

(D) $1 \, \text{mole}$ of glucose $= 180 \, \text{g}$ (molar mass of glucose).
$\text{Concentration } (\% \, w/v) = \frac{\text{Mass of solute (g)}}{\text{Volume of solution (mL)}} \times 100$
Given,$10 = \frac{180 \, \text{g}}{\text{Volume (mL)}} \times 100$
$\text{Volume (mL)} = \frac{180 \times 100}{10} = 1800 \, \text{mL}$
$\text{Volume (L)} = \frac{1800}{1000} = 1.8 \, \text{L}$

(D) The percentage strength of a solution can be expressed in different ways depending on the physical states of the solute and solvent:
$1$. Mass percentage $(w/w)$: $\frac{\text{Weight of solute}}{\text{Weight of solution}} \times 100$
$2$. Mass by volume percentage $(w/v)$: $\frac{\text{Weight of solute}}{\text{Volume of solution}} \times 100$
$3$. Volume percentage $(v/v)$: $\frac{\text{Volume of solute}}{\text{Volume of solution}} \times 100$
Since all these formulas represent valid ways to express the percentage concentration of a solution,the correct answer is $D$.

(C) The number of milliequivalents $(Meq.)$ in the concentrated solution is $1600 \times 0.2050 = 328$.
Let the final volume after dilution be $V \, mL$.
The number of milliequivalents in the diluted solution is $0.20 \times V$.
Since the number of milliequivalents remains constant during dilution,we have $328 = 0.20 \times V$.
Solving for $V$,we get $V = \frac{328}{0.20} = 1640 \, mL$.
The volume of water to be added is the difference between the final volume and the initial volume: $1640 \, mL - 1600 \, mL = 40 \, mL$.

(A) Given: $93\% \ (w/v)$ $H_2SO_4$ means $93 \ g$ of $H_2SO_4$ is present in $100 \ mL$ of solution.
Therefore,in $1000 \ mL$ $(1 \ L)$ of solution,the mass of $H_2SO_4$ is $930 \ g$.
Molar mass of $H_2SO_4 = 2 \times 1 + 32 + 4 \times 16 = 98 \ g/mol$.
Number of moles of $H_2SO_4 = \frac{\text{Mass}}{\text{Molar mass}} = \frac{930}{98} \approx 9.49 \ mol$.
Molarity $(M) = \frac{\text{Moles of solute}}{\text{Volume of solution in } L} = \frac{9.49}{1} = 9.49 \ M$.
Note: The provided options suggest the question might be asking for molality or there is a discrepancy in the provided options. Based on standard molarity calculation,the result is $9.49 \ M$. However,if we calculate molality $(m)$ using density $d = 1.84 \ g/mL$:
Mass of solution $= 1000 \ mL \times 1.84 \ g/mL = 1840 \ g$.
Mass of solvent $= 1840 \ g - 930 \ g = 910 \ g = 0.91 \ kg$.
Molality $(m) = \frac{9.49 \ mol}{0.91 \ kg} \approx 10.43 \ m$. Thus,the correct answer corresponds to the calculated molality.

(A) $25\%$ solution means $25 \ g$ of solute in $100 \ g$ of solution.
$40\%$ solution means $40 \ g$ of solute in $100 \ g$ of solution.
Mass of solute in $300 \ g$ solution $= \frac{25 \times 300}{100} = 75 \ g$.
Mass of solute in $400 \ g$ solution $= \frac{40 \times 400}{100} = 160 \ g$.
Total mass of solute $= 75 + 160 = 235 \ g$.
Total mass of solution $= 300 + 400 = 700 \ g$.
Percentage by weight $= \frac{235}{700} \times 100 = 33.57\%$.

(A) Molality $(m)$ is defined as the number of moles of solute present in $1 \ kg$ $(1000 \ g)$ of solvent.
Therefore,a $1$ molal solution contains $1$ mole of solute in $1000 \ g$ of solvent.

(A) Step $1$: Calculate the molar mass of benzene $(C_6H_6)$ and toluene $(C_6H_5CH_3)$.
$M(C_6H_6) = (6 \times 12) + (6 \times 1) = 78 \ g/mol$.
$M(C_6H_5CH_3) = (7 \times 12) + (8 \times 1) = 92 \ g/mol$.
Step $2$: Calculate the number of moles of each component.
$n(C_6H_6) = \frac{7.8 \ g}{78 \ g/mol} = 0.1 \ mol$.
$n(C_6H_5CH_3) = \frac{46.0 \ g}{92 \ g/mol} = 0.5 \ mol$.
Step $3$: Calculate the mole fraction of benzene $(x_{benzene})$.
$x_{benzene} = \frac{n(C_6H_6)}{n(C_6H_6) + n(C_6H_5CH_3)} = \frac{0.1}{0.1 + 0.5} = \frac{0.1}{0.6} = \frac{1}{6}$.

(B) $1 \ ppm$ (part per million) by mass means $1 \ part$ of solute in $10^6 \ parts$ of solution by mass.
Given,concentration of $CHCl_3 = 15 \ ppm$.
This means $15 \ g$ of $CHCl_3$ is present in $10^6 \ g$ of the water sample.
Percentage by mass $= \frac{\text{Mass of solute}}{\text{Mass of solution}} \times 100$.
Percentage by mass $= \frac{15}{10^6} \times 100 = 15 \times 10^{-4} = 1.5 \times 10^{-3} \%$.

(C) Mass of solute (urea) = $120 \, g$
Molar mass of urea = $60 \, g/mol$
Moles of urea = $\frac{120}{60} = 2 \, mol$
Mass of solvent (water) = $1000 \, g$
Total mass of solution = $120 \, g + 1000 \, g = 1120 \, g$
Density of solution = $1.15 \, g/mL$
Volume of solution = $\frac{\text{Mass}}{\text{Density}} = \frac{1120}{1.15} \, mL = \frac{1120}{1.15 \times 1000} \, L = \frac{1.12}{1.15} \, L$
Molarity $(M)$ = $\frac{\text{Moles of solute}}{\text{Volume of solution in } L} = \frac{2}{1.12 / 1.15} = \frac{2 \times 1.15}{1.12} \approx 2.05 \, M$

(B) Molarity $(M)$ is defined as the number of moles of solute per liter of solution.
Given:
Mass of $H_2O_2 = 5.0 \ g$
Volume of solution = $100 \ mL = 0.1 \ L$
Molar mass of $H_2O_2 = 34 \ g/mol$
Step $1$: Calculate the number of moles of $H_2O_2$.
Moles = $\frac{\text{Mass}}{\text{Molar mass}} = \frac{5.0 \ g}{34 \ g/mol} \approx 0.147 \ mol$
Step $2$: Calculate Molarity.
$M = \frac{\text{Moles of solute}}{\text{Volume of solution in liters}} = \frac{0.147 \ mol}{0.1 \ L} = 1.47 \ M \approx 1.5 \ M$.
Therefore,the correct option is $B$.

(C) To find the normality $(N)$ of each solution:
$A$: $8 \ g$ $KOH$ per liter. Molar mass of $KOH = 56 \ g/mol$. $N = \frac{8/56}{1} = 0.143 \ N$.
$B$: $1 \ N$ phosphoric acid is given as $1 \ N$.
$C$: $6 \ g$ $NaOH$ per $100 \ mL = 60 \ g$ per liter. Molar mass of $NaOH = 40 \ g/mol$. $N = \frac{60/40}{1} = 1.5 \ N$.
$D$: $0.5 \ M$ $H_2SO_4$. $N = M \times \text{valency factor} = 0.5 \times 2 = 1.0 \ N$.
Comparing the values: $0.143, 1, 1.5, 1.0$. The maximum is $1.5 \ N$.

(A) $1$. Calculate the number of moles of urea: $n = \frac{\text{mass}}{\text{molar mass}} = \frac{120 \ g}{60 \ g/mol} = 2 \ mol$.
$2$. Calculate the total mass of the solution: $\text{Mass of solution} = \text{mass of solute} + \text{mass of solvent} = 120 \ g + 1000 \ g = 1120 \ g$.
$3$. Calculate the volume of the solution: $V = \frac{\text{mass}}{\text{density}} = \frac{1120 \ g}{1.15 \ g/mL} \approx 973.91 \ mL = 0.97391 \ L$.
$4$. Calculate the molarity: $M = \frac{n}{V(L)} = \frac{2 \ mol}{0.97391 \ L} \approx 2.05 \ M$.

(B) Given: Concentration of $CHCl_3 = 15 \ ppm$ (by mass).
This means $15 \ g$ of $CHCl_3$ is present in $10^6 \ g$ of the solution.
Since the solution is dilute,we assume the density of the solution is approximately $1 \ g/mL$,so the volume of $10^6 \ g$ of solution is $10^6 \ mL = 1000 \ L$.
Molar mass of $CHCl_3 = 12 + 1 + 3 \times 35.5 = 119.5 \ g/mol$.
Number of moles of $CHCl_3 = \frac{15 \ g}{119.5 \ g/mol} \approx 0.1255 \ mol$.
Molarity $(M) = \frac{\text{moles of solute}}{\text{volume of solution in } L} = \frac{0.1255 \ mol}{1000 \ L} = 1.255 \times 10^{-4} \ M$.

(B) Molality $(X)$ is defined as the number of moles of solute per $1000 \ g$ of solvent.
Given,mass of solvent (benzene,$C_6H_6$) = $1000 \ g$.
Molar mass of benzene = $78 \ g/mol$.
Moles of benzene = $\frac{1000}{78} \approx 12.82 \ mol$.
Mole fraction of solute $(x_{solute})$ = $\frac{n_{solute}}{n_{solute} + n_{solvent}} = 0.2$.
Let $n_{solute} = X$.
$\frac{X}{X + 12.82} = 0.2$.
$X = 0.2X + 2.564$.
$0.8X = 2.564$.
$X = \frac{2.564}{0.8} = 3.205 \approx 3.2$.

(B) Concentration terms that involve volume (such as Molarity,Normality,and Gram per liter) are dependent on temperature because volume changes with temperature.
Molality $(m)$ is defined as the number of moles of solute per kilogram of solvent.
Since mass does not change with temperature,Molality is independent of temperature.

(D) Statement $(1)$ is incorrect because molarity is defined as moles of solute per liter of $solution$,not $solvent$.
Statement $(2)$ is incorrect because for $Na_2CO_3$,the $n$-factor is $2$,so $Normality = 2 \times Molarity$.
Statement $(3)$ is incorrect because $1000 \ g$ of solvent refers to molality $(m)$,but the statement labels it as molarity $(m)$,which is a notation error (molarity is $M$).
Statement $(4)$ is correct. The mole fraction $x_A = \frac{n_A}{n_A + n_B}$ and $x_B = \frac{n_B}{n_A + n_B}$. Therefore,the ratio $\frac{x_A}{x_B} = \frac{n_A}{n_B}$.

(C) The relationship between molarity $(M)$,density ($d$ in $g \ mL^{-1}$),and percentage by mass $(w/w\%)$ is given by the formula:
$M = \frac{w/w\% \times d \times 10}{Molar \ mass}$
Given:
$M = 3.60 \ M$
$w/w\% = 29\%$
$Molar \ mass = 98 \ g \ mol^{-1}$
Substituting the values:
$3.60 = \frac{29 \times d \times 10}{98}$
$3.60 = \frac{290 \times d}{98}$
$d = \frac{3.60 \times 98}{290}$
$d = \frac{352.8}{290} \approx 1.2165 \ g \ mL^{-1}$
Rounding to two decimal places,we get $d \approx 1.22 \ g \ mL^{-1}$.

(B) Molarity is defined as the number of moles of solute per liter of solution. $M = \frac{n}{V(L)}$.
As temperature increases,the volume of the solution $(V)$ increases due to thermal expansion.
Since $V$ is in the denominator,an increase in $V$ leads to a decrease in molarity $(M)$.

(D) The concentration is given as $10\% (W/V)$,which means $10 \ g$ of $H_2SO_4$ is present in $100 \ mL$ of the solution.
Normality $(N)$ is defined as the number of gram equivalents of solute per liter of solution.
$N = \frac{\text{Mass of solute in } g}{\text{Equivalent mass of solute} \times \text{Volume of solution in } L}$.
The molar mass of $H_2SO_4$ is $98 \ g/mol$. Since it is a dibasic acid,its n-factor is $2$.
Equivalent mass of $H_2SO_4 = \frac{98}{2} = 49 \ g/eq$.
Volume of solution = $100 \ mL = 0.1 \ L$.
$N = \frac{10 \ g}{49 \ g/eq \times 0.1 \ L} = \frac{10}{4.9} \approx 2.04 \ N$.
Therefore,the approximate normality is $2$.

(B) The molarity of the mixture is calculated using the formula: $M_3 = \frac{M_1V_1 + M_2V_2}{V_1 + V_2}$
Given:
$M_1 = 0.5 \ M$,$V_1 = 800 \ mL$
$M_2 = 1 \ M$,$V_2 = 200 \ mL$
Total volume $V_3 = V_1 + V_2 = 800 \ mL + 200 \ mL = 1000 \ mL$
Substituting the values:
$M_3 = \frac{(0.5 \times 800) + (1 \times 200)}{1000}$
$M_3 = \frac{400 + 200}{1000} = \frac{600}{1000} = 0.6 \ M$

(D) Mass of solute $(CH_3COOH)$ = $5 \ g$.
Molar mass of $CH_3COOH$ = $60 \ g/mol$.
Moles of solute = $\frac{5}{60} \approx 0.0833 \ mol$.
Volume of solvent (ethanol) = $1000 \ mL$.
Mass of solvent = $\text{Density} \times \text{Volume} = 0.789 \ g/mL \times 1000 \ mL = 789 \ g = 0.789 \ kg$.
Molality $(m)$ = $\frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.0833 \ mol}{0.789 \ kg} \approx 0.1056 \ m \approx 0.106 \ m$.

(C) The molarity $(M)$ of a pure liquid can be calculated using the formula: $M = \frac{d \times 1000}{M_w}$.
Here,$d$ is the density in $g/cc$ and $M_w$ is the molar mass of $HCl$.
The molar mass of $HCl = 1 + 35.5 = 36.5 \, g/mol$.
Given density $d = 1.17 \, g/cc$.
$M = \frac{1.17 \times 1000}{36.5} = \frac{1170}{36.5} \approx 32.05 \, M$.

(C) Molarity $(M)$ is defined as the number of moles of solute per liter of solution: $M = \frac{n}{V} = \frac{w}{MW \times V}$.
Since the weight $(w)$ and volume $(V)$ are the same for both,$M \propto \frac{1}{MW}$.
The molar mass of $NaCl$ is $23 + 35.5 = 58.5 \ g/mol$.
The molar mass of $KCl$ is $39 + 35.5 = 74.5 \ g/mol$.
Since the molar mass of $NaCl$ $(58.5 \ g/mol)$ is less than the molar mass of $KCl$ $(74.5 \ g/mol)$,the molarity of the $NaCl$ solution will be greater than that of the $KCl$ solution.

(A) Total mass of the solution $= \text{mass of aspirin} + \text{mass of water} = 3.65 \, g + 25.08 \, g = 28.73 \, g$.
Mass fraction of aspirin $= \frac{\text{mass of aspirin}}{\text{total mass of solution}} = \frac{3.65 \, g}{28.73 \, g} \approx 0.127$.
Mass percentage of aspirin $= \text{mass fraction} \times 100 = 0.127 \times 100 = 12.7\%$.

(C) $1 \, m$ (molal) solution means $1 \, \text{mole}$ of solute is dissolved in $1000 \, g$ of solvent (water).
Number of moles of solute $(n)$ = $1 \, \text{mol}$.
Mass of solvent $(W)$ = $1000 \, g$.
Molar mass of water $(H_2O)$ = $18 \, g/mol$.
Number of moles of solvent $(N)$ = $\frac{1000}{18} = 55.55 \, \text{mol}$.
Mole fraction of solute $(x)$ = $\frac{n}{n + N} = \frac{1}{1 + 55.55} = \frac{1}{56.55} \approx 0.0177 \approx 0.018$.

(C) Using the dilution formula: $M_1V_1 = M_2V_2$
Here,$M_1 = 3.0 \ M$,$M_2 = 1.0 \ M$,and $V_2 = 1.0 \ L = 1000 \ mL$.
Substituting the values: $3.0 \times V_1 = 1.0 \times 1000$
$V_1 = \frac{1000}{3} \ mL = 333.3 \ mL$.

(D) The normality of a component in a mixture is calculated by the formula $N_{final} = \frac{N_1V_1 + N_2V_2}{V_{total}}$.
Here,we are asked for the normality of $H_2SO_4$ specifically in the final mixture.
Initial volume of $H_2SO_4$ $(V_2)$ = $200 \ mL$.
Initial normality of $H_2SO_4$ $(N_2)$ = $0.6 \ N$.
Total volume of the mixture $(V_{total})$ = $100 \ mL + 200 \ mL = 300 \ mL$.
The amount of $H_2SO_4$ (in terms of milliequivalents) remains constant as it is not reacting with $HCl$.
$N_{final} \times V_{total} = N_2 \times V_2$.
$N_{final} = \frac{0.6 \ N \times 200 \ mL}{300 \ mL} = \frac{120}{300} \ N = 0.4 \ N$.