Method of expressing concentration of solution Questions in English

(A) The unit parts per million $(ppm)$ is used to express the concentration of a solution when the solute is present in trace quantities (very small amounts).
It is defined as the number of parts of solute per million parts of the solution by mass or volume.

(B) The unit $\% \ w/V$ stands for percentage weight by volume.
It is defined as the mass of the solute in grams dissolved in $100 \ mL$ of the solution.
This unit is commonly used in $Chemistry$ to express the concentration of solutions.

(A) Molarity $(M)$ is defined as the number of moles of solute dissolved in $1 \ L$ of solution.
It is expressed as:
$M = \frac{\text{Number of moles of solute}}{\text{Volume of solution in Liters}}$
The unit of molarity is $mol \ L^{-1}$ or $M$.

(A) The molar mass of $NaCl$ is $23.0 + 35.5 = 58.5 \ g/mol$.
Number of moles of $NaCl = \frac{\text{mass}}{\text{molar mass}} = \frac{29.22 \ g}{58.5 \ g/mol} = 0.4995 \ mol \approx 0.50 \ mol$.
Molality $(m)$ is defined as the number of moles of solute per kilogram of solvent.
$m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.50 \ mol}{2.00 \ kg} = 0.25 \ m$.

(A) Step $1$: Calculate the moles of solute $(NaCl)$.
Molar mass of $NaCl = 23 + 35.5 = 58.5 \ g/mol$.
Moles of $NaCl = \frac{2.89 \ g}{58.5 \ g/mol} \approx 0.0494 \ mol$.
Step $2$: Calculate the mass of the solvent (water).
Volume of water = $0.159 \ L = 159 \ mL$.
Density of water = $1.00 \ g/mL$.
Mass of water = $159 \ mL \times 1.00 \ g/mL = 159 \ g = 0.159 \ kg$.
Step $3$: Calculate molality $(m)$.
$m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.0494 \ mol}{0.159 \ kg} \approx 0.311 \ m$.

(N/A) majority of chemical reactions in laboratories are carried out in solutions. Therefore,it is important to understand 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 the following ways:
$(i)$ Mass percent or weight percent $(w/w \%)$
$(ii)$ Mole fraction
$(iii)$ Molarity
$(iv)$ Molality
Mass percent is defined as the mass of the solute per $100 \ g$ of the solution.
$\text{Mass percent} = \frac{\text{Mass of solute}}{\text{Mass of solution}} \times 100$

(A) Mass percentage of $A = \frac{\text{Mass of } A}{\text{Mass of solution}} \times 100$
Mass of solution $= \text{Mass of } A + \text{Mass of water} = 2 \ g + 18 \ g = 20 \ g$
Mass percentage of $A = \frac{2 \ g}{20 \ g} \times 100 = 10 \%$

(N/A) Mole fraction is defined as the ratio of the number of moles of a particular component to the total number of moles of all components present in the solution.
If a substance $A$ dissolves in substance $B$ and their number of moles are $n_{A}$ and $n_{B}$ respectively,then the mole fractions of $A$ and $B$ are given as:
Mole fraction of $A$ $(x_{A})$ = $\frac{n_{A}}{n_{A} + n_{B}}$
Mole fraction of $B$ $(x_{B})$ = $\frac{n_{B}}{n_{A} + n_{B}}$
The sum of mole fractions of all components in a solution is always equal to $1$ (i.e.,$x_{A} + x_{B} = 1$).

(C) The molar mass of $NaOH$ is $23 + 16 + 1 = 40 \ g/mol$.
Number of moles of $NaOH = \frac{\text{Given mass}}{\text{Molar mass}} = \frac{4 \ g}{40 \ g/mol} = 0.1 \ mol$.
The volume of the solution is $250 \ mL = 0.25 \ L$.
Molarity $(M) = \frac{\text{Number of moles of solute}}{\text{Volume of solution in } L} = \frac{0.1 \ mol}{0.25 \ L} = 0.4 \ M$.

(N/A) Molarity: It is the most widely used unit and is denoted by $M$.
It is defined as the number of moles of the solute in $1 \ L$ of the solution.
$\text{Molarity } (M) = \frac{\text{No. of moles of solute}}{\text{Volume of solution in litres}}$
Suppose we have $1 \ M$ solution of a substance,say $NaOH$,and we want to prepare a $0.2 \ M$ solution from it.
$1 \ M \ NaOH$ means $1 \ mol$ of $NaOH$ present in $1 \ L$ of the solution. For $0.2 \ M$ solution,we require $0.2 \ mol$ of $NaOH$ in $1 \ L$ solution.
For the calculation of molarity of two solutions,the following formula is used:
$M_1 \times V_1 = M_2 \times V_2$
Molality: It is defined as the number of moles of solute present in $1 \ kg$ of solvent. It is denoted by $m$.
$\text{Molality } (m) = \frac{\text{No. of moles of solute}}{\text{Mass of solvent in } kg}$

(A) Molarity is defined as the number of moles of solute per liter of solution.
Assuming $1 \ L$ of water,the number of moles of $H_2O$ is $n_{H_2O} = \frac{1000 \ g}{18 \ g/mol} = 55.55 \ mol$.
For a binary solution,the sum of mole fractions is $X_{H_2O} + X_{C_2H_5OH} = 1$.
Given $X_{C_2H_5OH} = 0.040$,then $X_{H_2O} = 1 - 0.040 = 0.96$.
The mole fraction formula is $X_{H_2O} = \frac{n_{H_2O}}{n_{H_2O} + n_{C_2H_5OH}}$.
Substituting the values: $0.96 = \frac{55.55}{55.55 + n_{C_2H_5OH}}$.
Solving for $n_{C_2H_5OH}$: $53.328 + 0.96 \times n_{C_2H_5OH} = 55.55$.
$0.96 \times n_{C_2H_5OH} = 2.222$.
$n_{C_2H_5OH} = \frac{2.222}{0.96} \approx 2.3145 \ mol$.
Since this is in $1 \ L$ of solution,the molarity is $2.31 \ M$.

(A) Given: Molarity $(M)$ = $3.0 \, mol \, L^{-1}$,Density $(d)$ = $1.25 \, g \, mL^{-1}$.
In $1 \, L$ of solution,the number of moles of $NaCl$ is $3 \, mol$.
Mass of $NaCl$ = $3 \, mol \times 58.5 \, g \, mol^{-1} = 175.5 \, g$.
Mass of $1 \, L$ solution = $Volume \times Density = 1000 \, mL \times 1.25 \, g \, mL^{-1} = 1250 \, g$.
Mass of solvent (water) = $Mass \, of \, solution - Mass \, of \, solute = 1250 \, g - 175.5 \, g = 1074.5 \, g = 1.0745 \, kg$.
Molality $(m)$ = $\frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{3 \, mol}{1.0745 \, kg} \approx 2.79 \, m$.

(A) The formula for molarity is: $M = \frac{w \times 1000}{m \times V(mL)}$
Where:
$w = \text{mass of solute in } g$
$m = \text{molar mass of solute} = 82.0245 \ g \ mol^{-1}$
$V = \text{volume of solution} = 500 \ mL$
$M = \text{molarity} = 0.375 \ M$
Rearranging for $w$:
$w = \frac{M \times m \times V(mL)}{1000}$
$w = \frac{0.375 \times 82.0245 \times 500}{1000}$
$w = 15.3796 \approx 15.38 \ g$

(A) The molar mass of sugar $(C_{12}H_{22}O_{11})$ is $12 \times 12 + 22 \times 1 + 11 \times 16 = 342 \ g \ mol^{-1}$.
The number of moles of sugar is $n = \frac{\text{mass}}{\text{molar mass}} = \frac{20 \ g}{342 \ g \ mol^{-1}} \approx 0.05848 \ mol$.
The molarity $(M)$ is defined as $\frac{\text{moles of solute}}{\text{volume of solution in } L}$.
$M = \frac{0.05848 \ mol}{2 \ L} = 0.02924 \ mol \ L^{-1}$.

(N/A) $(i)$ $15 \ ppm$ means $15$ parts per million. Therefore,percent by mass $= \frac{15 \times 100}{10^6} = 1.5 \times 10^{-3} \%$.
$(ii)$ Molar mass of $CHCl_3 = 12.01 + 1.008 + (3 \times 35.45) = 119.37 \ \text{g/mol}$.
$1.5 \times 10^{-3} \%$ means $1.5 \times 10^{-3} \ \text{g}$ of $CHCl_3$ is present in $100 \ \text{g}$ of solution.
Mass of solvent (water) $\approx 100 \ \text{g} = 0.1 \ \text{kg}$.
$\text{Molality} (m) = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{1.5 \times 10^{-3} \ \text{g} / 119.37 \ \text{g/mol}}{0.1 \ \text{kg}} = 1.256 \times 10^{-4} \ \text{m} \approx 1.26 \times 10^{-4} \ \text{m}$.

(B) $Molality$: It is defined as the number of moles of solute dissolved in $1 \ kg$ of solvent. It is independent of temperature.
$Molarity$: It is defined as the number of moles of solute dissolved in $1 \ L$ of solution. It depends upon temperature.

(A) $3 \ m$ solution of $NaOH$ means $3 \ moles$ of $NaOH$ are dissolved in $1000 \ g$ of solvent.
Molar mass of $NaOH = 23 + 16 + 1 = 40 \ g \ mol^{-1}$.
Mass of $NaOH = 3 \ moles \times 40 \ g \ mol^{-1} = 120 \ g$.
Total mass of solution = Mass of solvent + Mass of solute = $1000 \ g + 120 \ g = 1120 \ g$.
Volume of solution = $\frac{\text{Mass}}{\text{Density}} = \frac{1120 \ g}{1.110 \ g \ mL^{-1}} \approx 1009.01 \ mL$.
Molarity $(M)$ = $\frac{\text{Moles of solute} \times 1000}{\text{Volume of solution in } mL} = \frac{3 \times 1000}{1009.01} \approx 2.97 \ M$.

(N/A) The molality of a solution is not affected by a change in temperature.
$Molality (m) = \frac{\text{moles of solute}}{\text{mass of solvent (in kg)}}$.
Since the mass of the solute and the mass of the solvent remain constant regardless of temperature changes,the molality remains unchanged.

Mass of $NaOH = 4 \ g$
Number of moles of $NaOH = \frac{4 \ g}{40 \ g \ mol^{-1}} = 0.1 \ mol$
Mass of $H_2O = 36 \ g$
Number of moles of $H_2O = \frac{36 \ g}{18 \ g \ mol^{-1}} = 2 \ mol$
Mole fraction of water $(x_{H_2O})$ $= \frac{2}{2 + 0.1} = \frac{2}{2.1} \approx 0.952$
Mole fraction of $NaOH$ $(x_{NaOH})$ $= \frac{0.1}{2 + 0.1} = \frac{0.1}{2.1} \approx 0.048$
Mass of solution $= 36 \ g + 4 \ g = 40 \ g$
Volume of solution $= \frac{\text{Mass}}{\text{Density}} = \frac{40 \ g}{1 \ g \ mL^{-1}} = 40 \ mL = 0.04 \ L$
Molarity of solution $= \frac{\text{moles of solute}}{\text{Volume of solution in } L} = \frac{0.1 \ mol}{0.04 \ L} = 2.5 \ M$

(N/A) Molarity of a solution is defined as the number of moles of solute dissolved in $1 \ L$ of solution. Since volume depends on temperature and changes with it,molarity also changes with temperature.
On the other hand,concentration terms like mass percentage,$ppm$,mole fraction,and molality are based on mass-by-mass relationships.
Since mass does not change with temperature,these concentration terms remain independent of temperature variations.

(N/A) $1$. Mass percentage: The mass of the solute in grams present in $100 \ g$ of the solution.
$2$. Molarity $(M)$: The number of moles of solute dissolved in $1 \ L$ of solution.
$3$. Molality $(m)$: The number of moles of solute dissolved in $1 \ kg$ of solvent.
$4$. Mole fraction $(x)$: The ratio of the number of moles of one component to the total number of moles of all components in the solution.
Modes independent of temperature are Molality and Mole fraction.
Reason: These modes are defined in terms of mass (or moles),which do not change with temperature. In contrast,Molarity involves volume,which changes with temperature due to thermal expansion or contraction.

(B) The density of pure water is approximately $1.0 \ g \ mL^{-1}$ or $1000 \ g \ L^{-1}$.
The molar mass of water $(H_2O)$ is $2 \times 1.008 + 16.00 = 18.016 \ g \ mol^{-1}$,which is approximately $18 \ g \ mol^{-1}$.
Concentration (Molarity) $= \frac{\text{Mass of solute in } g \ L^{-1}}{\text{Molar mass of solute in } g \ mol^{-1}}$.
Concentration of water $= \frac{1000 \ g \ L^{-1}}{18 \ g \ mol^{-1}} \approx 55.5 \ mol \ L^{-1}$.
Therefore,statement $(a)$ is False $(F)$ and statement $(b)$ is True $(T)$.

(A) The concentration of a solution is defined as the amount of solute present in a given quantity (mass or volume) of the solution or solvent.
Mathematically,it is expressed as: $\text{Concentration} = \frac{\text{Amount of solute}}{\text{Amount of solution}}$.

(N/A) The amount of solute dissolved in a unit volume or unit mass of the solution is called the concentration of the solution.
Some common methods to express the concentration of a solution are as follows:
$(i)$ Molarity $(M)$
$(ii)$ Molality $(m)$
$(iii)$ Normality $(N)$
$(iv)$ Mole fraction $(x)$
$(v)$ Mass percentage $(\%w/w)$
$(vi)$ Volume percentage $(\%v/v)$
$(vii)$ Mass by volume percentage $(\%w/v)$

(B) Mass percentage $(\% w/w)$ is defined as the mass of the solute in grams present in $100 \ g$ of the solution.
Formula: $\% w/w = \frac{\text{Mass of solute}}{\text{Mass of solution}} \times 100$

(B) The sum of the mole fractions of all the components in a solution is always equal to $1$.
Mathematically,for a solution with components $A, B, C, ...$,the sum is $\sum x_i = x_A + x_B + x_C + ... = 1$.

(B) The molarity $(M)$ of $H_2SO_4$ is $5 \ M$. The n-factor for $H_2SO_4$ is $2$.
Normality $(N)$ = Molarity $(M)$ $\times$ n-factor = $5 \times 2 = 10 \ N$.
Using the dilution formula $N_1V_1 = N_2V_2$:
$10 \ N \times 1 \ L = N_2 \times 10 \ L$.
$N_2 = \frac{10 \times 1}{10} = 1 \ N$.

(A) Given that the mixture contains $54 \%$ water by mass,we assume $100 \ g$ of the mixture.
Mass of $H_2O = 54 \ g$.
Mass of ethanol $(C_2H_5OH)$ $= 100 \ g - 54 \ g = 46 \ g$.
Moles of $H_2O = \frac{54 \ g}{18 \ g/mol} = 3 \ mol$.
Moles of ethanol $= \frac{46 \ g}{46 \ g/mol} = 1 \ mol$.
Total moles $= 3 + 1 = 4 \ mol$.
Mole fraction of ethanol $= \frac{\text{moles of ethanol}}{\text{total moles}} = \frac{1}{4} = 0.25$.

(B) Given,mole fraction of glucose $X_{C_{6}H_{12}O_{6}} = 0.1$.
Since it is a binary solution,the mole fraction of water $X_{H_{2}O} = 1 - 0.1 = 0.9$.
Let the total number of moles be $1 \ mol$.
Then,moles of glucose $n_{glucose} = 0.1 \ mol$ and moles of water $n_{water} = 0.9 \ mol$.
Mass of glucose $= 0.1 \ mol \times 180 \ g/mol = 18 \ g$.
Mass of water $= 0.9 \ mol \times 18 \ g/mol = 16.2 \ g$.
Total mass of the solution $= 18 \ g + 16.2 \ g = 34.2 \ g$.
Mass percentage of water $= (\text{Mass of water} / \text{Total mass of solution}) \times 100 = (16.2 / 34.2) \times 100 \approx 47.36\%$.
Rounding to the nearest integer,we get $47$.

(D) Molarity $(C_{2})$ is defined as the number of moles of solute per liter of solution.
$C_{2} = \frac{n_{2}}{V_{sol} (in \ L)} = \frac{n_{2} \times 1000}{V_{sol} (in \ mL)}$
Since $V_{sol} = \frac{\text{Total Mass}}{d} = \frac{n_{1}M_{1} + n_{2}M_{2}}{d}$, we have:
$C_{2} = \frac{n_{2} \times 1000 \times d}{n_{1}M_{1} + n_{2}M_{2}}$
Divide numerator and denominator by $(n_{1} + n_{2})$:
$C_{2} = \frac{1000 d (n_{2} / (n_{1} + n_{2}))}{(n_{1}M_{1} + n_{2}M_{2}) / (n_{1} + n_{2})}$
Using mole fraction $x_{2} = \frac{n_{2}}{n_{1} + n_{2}}$ and $x_{1} = \frac{n_{1}}{n_{1} + n_{2}} = 1 - x_{2}$:
$C_{2} = \frac{1000 d x_{2}}{x_{1}M_{1} + x_{2}M_{2}} = \frac{1000 d x_{2}}{(1 - x_{2})M_{1} + x_{2}M_{2}}$
$C_{2} = \frac{1000 d x_{2}}{M_{1} - x_{2}M_{1} + x_{2}M_{2}} = \frac{1000 d x_{2}}{M_{1} + x_{2}(M_{2} - M_{1})}$

(A) The molarity $(M)$ of a solution is given by the formula: $M = \frac{\% \ w/w \times d \times 10}{M_{solute}}$.
Here,the percentage by weight $(\% \ w/w)$ is $63$,the density $(d)$ is $5.4 \ g/mL$,and the molar mass of $HNO_3$ $(M_{solute})$ is $1 + 14 + (3 \times 16) = 63 \ g/mol$.
Substituting these values into the formula:
$M = \frac{63 \times 5.4 \times 10}{63} = 54 \ M$.

(C) Molality $(m)$ is defined as the number of moles of solute per $1000 \ g$ of solvent.
Given $6.50$ molal $KOH$ solution means $6.50 \ moles$ of $KOH$ are present in $1000 \ g$ of water.
Molar mass of $KOH = 39.0 + 16.0 + 1.0 = 56.0 \ g \ mol^{-1}$.
Mass of solute $(KOH)$ $= 6.50 \ mol \times 56.0 \ g \ mol^{-1} = 364.0 \ g$.
Total mass of solution $= \text{Mass of solvent} + \text{Mass of solute} = 1000 \ g + 364.0 \ g = 1364.0 \ g$.
Density of solution $= 1.89 \ g \ cm^{-3}$.
Volume of solution $= \frac{\text{Mass}}{\text{Density}} = \frac{1364.0 \ g}{1.89 \ g \ cm^{-3}} \approx 721.69 \ cm^3 = 0.72169 \ L$.
Molarity $(M)$ $= \frac{\text{moles of solute}}{\text{Volume of solution in } L} = \frac{6.50 \ mol}{0.72169 \ L} \approx 9.006 \ mol \ L^{-1}$.
Rounding to the nearest integer,the molarity is $9 \ mol \ dm^{-3}$.

(A) $100$ molal aqueous solution means there are $100$ moles of solute in $1 \ kg$ $(1000 \ g)$ of water.
The number of moles of water $(n_{\text{solvent}})$ $= \frac{1000 \ g}{18 \ g/mol} = 55.56 \ mol$.
The mole fraction of solute $(x_{\text{solute}})$ $= \frac{n_{\text{solute}}}{n_{\text{solute}} + n_{\text{solvent}}}$.
$x_{\text{solute}} = \frac{100}{100 + 55.56} = \frac{100}{155.56} \approx 0.6428$.
$0.6428 = 64.28 \times 10^{-2}$.
Rounding off to the nearest integer,we get $64 \times 10^{-2}$.

(B) The number of moles of compound $A$ is calculated as: $n = \frac{\text{mass}}{\text{molar mass}} = \frac{4.5 \ g}{90 \ g/mol} = 0.05 \ mol$.
The volume of the solution in liters is: $V = 250 \ mL = 0.250 \ L$.
The molarity $(M)$ is given by: $M = \frac{n}{V} = \frac{0.05 \ mol}{0.250 \ L} = 0.2 \ M$.
Expressing this in the form $x \times 10^{-1}$: $0.2 = 2 \times 10^{-1}$.
Therefore,the value of $x$ is $2$.

(A) Given: Molality $(m)$ = $3.30 \ mol \ kg^{-1}$,Density $(d)$ = $1.20 \ g \ mL^{-1}$,Molar mass of $KCl$ $(M_2)$ = $74.5 \ g \ mol^{-1}$.
Let the mass of solvent be $1000 \ g$ $(1 \ kg)$.
Moles of solute $(n_2)$ = $3.30 \ mol$.
Mass of solute $(w_2)$ = $n_2 \times M_2 = 3.30 \times 74.5 = 245.85 \ g$.
Total mass of solution $(w)$ = Mass of solvent + Mass of solute = $1000 + 245.85 = 1245.85 \ g$.
Volume of solution $(V)$ = $\frac{\text{Mass}}{\text{Density}} = \frac{1245.85 \ g}{1.20 \ g \ mL^{-1}} = 1038.21 \ mL = 1.03821 \ L$.
Molarity $(M)$ = $\frac{\text{Moles of solute}}{\text{Volume of solution in } L} = \frac{3.30 \ mol}{1.03821 \ L} \approx 3.178 \ mol \ L^{-1}$.
The nearest integer is $3$.

(D) $1$. Calculate the molar mass of oxalic acid dihydrate $(H_{2}C_{2}O_{4} \cdot 2H_{2}O)$:
$M_w = (2 \times 1.0) + (2 \times 12.0) + (4 \times 16.0) + 2 \times (2 \times 1.0 + 16.0) = 2 + 24 + 64 + 36 = 126 \ g \ mol^{-1}$.
$2$. Calculate the number of moles of oxalic acid:
$n = \frac{\text{mass}}{M_w} = \frac{6.3 \ g}{126 \ g \ mol^{-1}} = 0.05 \ mol$.
$3$. Calculate the molarity $(M)$ of the solution:
$M = \frac{n}{V(L)} = \frac{0.05 \ mol}{0.250 \ L} = 0.2 \ mol \ L^{-1}$.
$4$. Express the molarity in the form $x \times 10^{-2}$:
$0.2 = 20 \times 10^{-2}$.
Therefore,the value of $x$ is $20$.

(C) $1$. Calculate the molar mass of $CuSO_{4} \cdot 5H_{2}O$:
$M = 63.54 + 32 + (4 \times 16) + 5 \times (2 \times 1 + 16) = 63.54 + 32 + 64 + 90 = 249.54 \ g \ mol^{-1}$.
$2$. Calculate the number of moles of $CuSO_{4} \cdot 5H_{2}O$:
$n = \frac{\text{mass}}{\text{molar mass}} = \frac{80 \ g}{249.54 \ g \ mol^{-1}} \approx 0.3206 \ mol$.
$3$. Calculate the molarity of the solution:
$Molarity = \frac{n}{V(L)} = \frac{0.3206 \ mol}{5 \ L} = 0.06412 \ mol \ L^{-1}$.
$4$. Express in the form $x \times 10^{-3} \ mol \ L^{-1}$:
$0.06412 = 64.12 \times 10^{-3}$.
Thus,the value of $x$ is $64$.

(A) The molar mass of glucose $(C_{6}H_{12}O_{6})$ is calculated as: $6 \times 12 + 12 \times 1 + 6 \times 16 = 72 + 12 + 96 = 180 \ g \ mol^{-1}$.
Molarity $(M)$ is defined as the number of moles of solute per liter of solution.
$M = \frac{\text{mass of solute (g)}}{\text{molar mass (g/mol)} \times \text{volume of solution (L)}}$.
Given concentration $= 0.72 \ g \ L^{-1}$,so for $1 \ L$ of solution,mass $= 0.72 \ g$.
$M = \frac{0.72 \ g}{180 \ g \ mol^{-1} \times 1 \ L} = 0.004 \ mol \ L^{-1}$.
$0.004 \ mol \ L^{-1} = 4 \times 10^{-3} \ M$.

(B) Assume $1 \, L$ of the solution.
Mass of solution $= \text{density} \times \text{volume} = 1.2 \, g \, cm^{-3} \times 1000 \, cm^3 = 1200 \, g$.
Assuming the molarity is $M$ (if not given,we typically assume a $1 \, M$ solution for such problems,or if the question implies $1 \, M$ based on the calculation steps provided in the original prompt).
If we assume $1 \, M$ $NaOH$ solution:
Moles of $NaOH = 1 \, mol$.
Mass of $NaOH = 1 \, mol \times 40 \, g \, mol^{-1} = 40 \, g$.
Mass of solvent (water) $= 1200 \, g - 40 \, g = 1160 \, g = 1.16 \, kg$.
Molality $(m) = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{1 \, mol}{1.16 \, kg} \approx 0.86 \, m$.
However,based on the provided solution steps in the prompt which calculate $5 \, mol$ of $NaOH$ in $1 \, kg$ of water,the question implies a $5 \, M$ solution.
For a $5 \, M$ solution:
Moles of $NaOH = 5 \, mol$.
Mass of $NaOH = 5 \, mol \times 40 \, g \, mol^{-1} = 200 \, g$.
Mass of solution $= 1200 \, g$.
Mass of solvent $= 1200 \, g - 200 \, g = 1000 \, g = 1 \, kg$.
Molality $(m) = \frac{5 \, mol}{1 \, kg} = 5 \, m$.

(A) Molality $(m)$ is defined as the number of moles of solute per kilogram of solvent.
$m = \frac{\text{Moles of solute}}{\text{Mass of solvent in } kg}$
Given $m = 1 \ mol/kg$ and $\text{moles of solute} = 0.5 \ mol$.
$1 = \frac{0.5}{\text{Mass of solvent in } kg}$
$\text{Mass of solvent in } kg = 0.5 \ kg = 500 \ g$.

(C) $1$. Given: Density $(d)$ = $1.46 \ g/mL$,Percentage by mass = $35 \%$,Molar mass of $HCl$ $(M_w)$ = $1 + 35.5 = 36.5 \ g/mol$.
$2$. Formula for Molarity $(M)$: $M = \frac{\text{Percentage by mass} \times d \times 10}{M_w}$.
$3$. Substituting the values: $M = \frac{35 \times 1.46 \times 10}{36.5}$.
$4$. Calculation: $M = \frac{35 \times 14.6}{36.5} = \frac{511}{36.5} = 14 \ M$.

(A) Molality is defined as the number of moles of solute per kilogram of solvent $(mol \ kg^{-1})$.
Since both the number of moles of solute and the mass of the solvent are independent of temperature,molality does not change with temperature.
Therefore,Assertion $A$ is true.
Reason $R$ correctly explains that mass remains unaffected by temperature,which is the reason why molality is temperature-independent.
Thus,both $A$ and $R$ are true and $R$ is the correct explanation of $A$.

(B) The molecular formula of glucose is $C_6H_{12}O_6$. The molar mass of glucose is $6 \times 12 + 12 \times 1 + 6 \times 16 = 180 \ g/mol$.
The mass of carbon in $180 \ g$ of glucose is $6 \times 12 = 72 \ g$.
Given that the solution contains $10.8 \ \%$ carbon by weight,the mass of carbon in $250 \ g$ of solution is $\frac{10.8}{100} \times 250 = 27 \ g$.
Since $72 \ g$ of carbon is present in $180 \ g$ of glucose,the mass of glucose containing $27 \ g$ of carbon is $\frac{180}{72} \times 27 = 67.5 \ g$.
The mass of the solvent (water) is $250 \ g - 67.5 \ g = 182.5 \ g = 0.1825 \ kg$.
The number of moles of glucose is $\frac{67.5 \ g}{180 \ g/mol} = 0.375 \ mol$.
Molality $(m) = \frac{\text{moles of solute}}{\text{mass of solvent in } kg} = \frac{0.375 \ mol}{0.1825 \ kg} \approx 2.055 \ m \approx 2.06 \ m$.

(C) Step $1$: Calculate the mass of the solution. $\text{Mass of solution} = \text{Volume} \times \text{Density} = 250\, mL \times 1.9\, g/mL = 475\, g$.
Step $2$: Calculate the mass of the solvent. $\text{Mass of solvent} = \text{Mass of solution} - \text{Mass of solute} = 475\, g - 3.0\, g = 472\, g$.
Step $3$: Calculate molality. $\text{Molar mass of } (CO_2H)_2 \cdot 2H_2O = 126\, g\, mol^{-1}$. $\text{Molality} = \frac{3.0 \times 1000}{126 \times 472} \approx 0.05\, mol\, kg^{-1}$.
Step $4$: Calculate normality. $\text{Equivalent mass of oxalic acid} = \frac{126}{2} = 63\, g/\text{equiv}$. $\text{Normality} = \frac{\text{Mass of solute} \times 1000}{\text{Equivalent mass} \times \text{Volume of solution } (mL)} = \frac{3.0 \times 1000}{63 \times 250} = 0.19\, N$.

(B) Given,weight percent of sucrose $= 3.42 \%$. This means $3.42 \, g$ of sucrose is present in $100 \, g$ of solution.
Molar mass of sucrose $= 342 \, g \, mol^{-1}$.
Number of moles of sucrose $= \frac{\text{mass}}{\text{molar mass}} = \frac{3.42 \, g}{342 \, g \, mol^{-1}} = 0.01 \, mol$.
Density of solution $= 1 \, g \, mL^{-1}$.
Volume of solution $= \frac{\text{mass of solution}}{\text{density}} = \frac{100 \, g}{1 \, g \, mL^{-1}} = 100 \, mL = 0.1 \, L$.
Molarity $= \frac{\text{moles of solute}}{\text{volume of solution in } L} = \frac{0.01 \, mol}{0.1 \, L} = 0.1 \, mol \, L^{-1}$.
Thus,the correct option is $B$.

(C) The molarity $(M)$ of a solution is defined as the number of moles of solute per liter of solution.
Given:
Number of moles of solute $(NaCl)$ = $0.35 \, \text{mol}$
Volume of solution = $1.30 \, \text{L}$
Calculation:
$M = \frac{\text{Number of moles of solute}}{\text{Volume of solution in liters}}$
$M = \frac{0.35}{1.30} \approx 0.26923 \, \text{M}$
Rounding to two decimal places,we get $0.27 \, \text{M}$.

(A) The units used to express the concentration of solutions are:
$1$. Mass percent
$2$. Mole fraction
$3$. Molarity
$4$. ppm (parts per million)
$5$. Molality
'Mole' is a unit of amount of substance,not a concentration unit.
Therefore,there are $5$ such units.

(C) Case $I$: For $0.25 m$ solution,molality $(m) = \frac{W_1 \times 1000}{M_2 \times W_{\text{solvent}}}$.
Assuming the mass of the solvent is approximately equal to the mass of the solution $(500 g)$:
$0.25 = \frac{W_1 \times 1000}{62 \times 500} \implies W_1 = \frac{0.25 \times 62}{2} = 7.75 g$.
Case $II$: For $0.25 M$ solution,molarity $(M) = \frac{W_2 \times 1000}{M_2 \times V_{\text{solution}}}$.
$0.25 = \frac{W_2 \times 1000}{62 \times 250} \implies W_2 = \frac{0.25 \times 62}{4} = 3.875 g$.
Ratio $\frac{W_1}{W_2} = \frac{7.75}{3.875} = \frac{2}{1}$.

(B) $1$. Calculate the molar mass of $CH_2Cl_2$: $12 + 2(1) + 2(35.5) = 85 \ g \ mol^{-1}$.
$2$. Calculate the moles of $CH_2Cl_2$ in $0.671141 \ L$ of solution: $\text{moles} = \text{Molarity} \times \text{Volume} = 2.6 \times 10^{-3} \ mol \ L^{-1} \times 0.671141 \ L = 1.745 \times 10^{-3} \ mol$.
$3$. Calculate the mass of $CH_2Cl_2$ $(x)$: $x = \text{moles} \times \text{molar mass} = 1.745 \times 10^{-3} \ mol \times 85 \ g \ mol^{-1} = 0.1483 \ g$.
$4$. Calculate the mass of the solvent $(CHCl_3)$: $\text{mass} = \text{density} \times \text{volume} = 1.49 \ g \ cm^{-3} \times 671.141 \ cm^3 = 1000 \ g$.
$5$. Calculate the concentration in $ppm$: $\text{ppm} = \frac{\text{mass of solute}}{\text{mass of solvent}} \times 10^6 = \frac{0.1483 \ g}{1000 \ g} \times 10^6 = 148.3 \ ppm \approx 148 \ ppm$.

(A) The molarity $(M)$ is $3 \ M$ and density $(d)$ is $1.0 \ g \ mL^{-1}$.
The molar mass of $NaCl$ is $23 + 35.5 = 58.5 \ g \ mol^{-1}$.
The formula for molality $(m)$ is $m = \frac{1000 \times M}{1000 \times d - M \times M_{solute}}$.
Substituting the values: $m = \frac{1000 \times 3}{1000 \times 1 - 3 \times 58.5} = \frac{3000}{1000 - 175.5} = \frac{3000}{824.5} \approx 3.6385 \ m$.
Expressing in $10^{-2} \ m$: $3.6385 \times 10^2 \times 10^{-2} \approx 364 \times 10^{-2} \ m$.
The nearest integer is $364$.