Method of expressing concentration of solution Questions in English

(A) Given: Mole fraction of solute $(x_2)$ = $0.2$.
Since the sum of mole fractions is $1$,the mole fraction of solvent (benzene,$C_6H_6$) is $x_1 = 1 - 0.2 = 0.8$.
The molar mass of benzene $(C_6H_6)$ is $M_1 = (6 \times 12) + (6 \times 1) = 78 \ g/mol$.
Molality $(m)$ is given by the formula: $m = \frac{x_2 \times 1000}{x_1 \times M_1}$.
Substituting the values: $m = \frac{0.2 \times 1000}{0.8 \times 78} = \frac{200}{62.4} \approx 3.205 \ m$.
Rounding to the nearest given option,the answer is $3.2$.

(A) Given:
Volume of solution $(V)$ = $1 \ L = 1000 \ mL$
Density of solution $(d)$ = $1.2 \ g/mL$
Concentration by weight = $40\%$
Step $1$: Calculate the total mass of the solution.
Mass of solution = $\text{Density} \times \text{Volume} = 1.2 \ g/mL \times 1000 \ mL = 1200 \ g$
Step $2$: Calculate the mass of the solute.
Mass of solute = $40\%$ of total mass of solution
Mass of solute = $\frac{40}{100} \times 1200 \ g = 480 \ g$
Therefore,the mass of the solute is $480 \ g$.

(A) The formula for concentration in parts per million $(ppm)$ is given by:
$ppm = \frac{\text{mass of solute}}{\text{mass of solution}} \times 10^6$
Given:
Mass of solute $(BaCl_2)$ = $1.04 \ g$
Mass of solvent (water) = $10^5 \ g$
Since the mass of solute is very small compared to the mass of solvent,the mass of the solution is approximately equal to the mass of the solvent:
Mass of solution $\approx 10^5 \ g$
Now,substituting the values:
$ppm = \frac{1.04}{10^5} \times 10^6$
$ppm = 1.04 \times 10^1 = 10.4$
Therefore,the concentration is $10.4 \ ppm$.

(B) Concentration terms that involve volume (like Molarity,Normality,and Gram/Liter) are temperature-dependent because volume changes with temperature.
Molality 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.

(B) The molarity $(M)$ of $Ca^{2+}$ is $0.0002 \ mol/L$.
To convert molarity to $ppm$ $(mg/L)$,we use the formula: $ppm = \text{Molarity} \times \text{Molar Mass} \times 1000$.
The molar mass of $Ca^{2+}$ is approximately $40 \ g/mol$.
$ppm = 0.0002 \ mol/L \times 40 \ g/mol \times 1000 \ mg/g$.
$ppm = 0.0002 \times 40000 = 8 \ ppm$.

(A) The concentration in $ppm$ (parts per million) is defined as the mass of solute in $mg$ per liter of solution.
Given,concentration = $10 \ ppm = 10 \ mg/L$.
We know that $1 \ mg = 10^{-3} \ g$.
So,$10 \ mg = 10 \times 10^{-3} \ g = 10^{-2} \ g$.
Thus,the concentration is $10^{-2} \ g/L$.
To convert this to $\% W/V$ (grams per $100 \ mL$):
Concentration in $\% W/V = \frac{\text{mass of solute in } g}{\text{volume of solution in } mL} \times 100$.
Concentration = $\frac{10^{-2} \ g}{1000 \ mL} \times 100 = \frac{10^{-2}}{10} = 10^{-3} \% W/V$.

(C) Given that molarity $(M)$ = molality $(m)$.
Formula for molarity: $M = \frac{n_2 \times 1000}{V_{sol} (mL)}$
Formula for molality: $m = \frac{n_2 \times 1000}{W_1 (g)}$
Since $M = m$,it implies $V_{sol} (mL) = W_1 (g)$,where $W_1$ is the mass of the solvent (water).
The mass of the solution $(W_{sol})$ = $W_1 + W_2$,where $W_2$ is the mass of urea ($NH_2CONH_2$,molar mass = $60 \ g/mol$).
$W_2 = M \times 60 \times \frac{V_{sol}}{1000} = \frac{60M \times V_{sol}}{1000} = 0.06M \times V_{sol}$.
Density $(d)$ = $\frac{W_{sol}}{V_{sol}} = \frac{W_1 + W_2}{V_{sol}} = \frac{V_{sol} + 0.06M \times V_{sol}}{V_{sol}} = 1 + 0.06M$.
$d = 1 + \frac{6M}{100} = 1 + \frac{3M}{50} = \frac{50 + 3M}{50}$.

(C) Given: Molarity $(M)$ = $1.56 \ M$,Molality $(m)$ = $1.8 \ m$.
The molar mass of $H_2SO_4$ $(M_2)$ = $(2 \times 1) + 32 + (4 \times 16) = 98 \ g/mol$.
The relationship between molarity and molality is given by the formula:
$d = M \times (\frac{1}{m} + \frac{M_2}{1000})$
Substituting the values:
$d = 1.56 \times (\frac{1}{1.8} + \frac{98}{1000})$
$d = 1.56 \times (0.5556 + 0.098)$
$d = 1.56 \times 0.6536 = 1.0196 \approx 1.02 \ g/mL$.

(B) Hardness in $ppm$ is defined as the mass of solute in $mg$ per $10^6 \, mg$ of water.
Since $1 \, kg$ of water is equal to $1000 \, g$ or $10^6 \, mg$,the concentration in $ppm$ is numerically equal to the mass of the solute in $mg$ per $1 \, kg$ of water.
Given that the hardness is $50 \, ppm$,it means there are $50 \, mg$ of $MgSO_4$ per $10^6 \, mg$ of water.
Therefore,in $1 \, kg$ of water,there are $50 \, mg$ of $MgSO_4$ present.
Thus,the correct option is $B$.

(B) Given mass of $NaOH$ $(W_{NaOH})$ = $24.5 \ g$
Molar mass of $NaOH$ $(M_{NaOH})$ = $40.0 \ g \ mol^{-1}$
Number of moles of $NaOH$ $(n)$ = $\frac{W_{NaOH}}{M_{NaOH}} = \frac{24.5}{40.0} = 0.6125 \ mol$
Volume of solution $(V)$ = $1 \ L$
Molarity $(M)$ = $\frac{n}{V(L)} = \frac{0.6125 \ mol}{1 \ L} = 0.6125 \ M$

(A) $Molality = \frac{\text{Number of moles of solute}}{\text{Weight of solvent in kg}}$
For glucose $(C_6H_{12}O_6)$,the molar mass is $180 \, g/mol$.
Number of moles of glucose = $\frac{180 \, g}{180 \, g/mol} = 1 \, mol$.
Since the solvent weight is $1 \, kg$,the molality is $\frac{1 \, mol}{1 \, kg} = 1 \, m$.
Thus,the assertion is correct.
The definition of a one molal solution is a solution containing $1 \, mole$ of solute in $1000 \, g$ $(1 \, kg)$ of solvent,which makes the reason correct and the correct explanation for the assertion.

(B) The mass of $O_2$ is $10.30 \; mg = 10.30 \times 10^{-3} \; g$.
The volume of seawater is $1 \; L = 1000 \; mL$.
The density of seawater is $1.03 \; g/mL$.
The mass of seawater = $\text{density} \times \text{volume} = 1.03 \; g/mL \times 1000 \; mL = 1030 \; g$.
The concentration in $ppm$ is calculated as:
$ppm = \frac{\text{mass of solute}}{\text{mass of solution}} \times 10^{6}$
$ppm = \frac{10.30 \times 10^{-3} \; g}{1030 \; g} \times 10^{6} = \frac{10.30}{1030} \times 10^{3} = 0.01 \times 1000 = 10$.

(A) Mass per cent of solute $(A)$ is calculated using the formula:
$\text{Mass per cent} = \frac{\text{Mass of solute}}{\text{Mass of solution}} \times 100$
Given:
Mass of solute $(A)$ = $2 \ g$
Mass of solvent (water) = $18 \ g$
Mass of solution = $\text{Mass of solute} + \text{Mass of solvent} = 2 \ g + 18 \ g = 20 \ g$
$\text{Mass per cent of } A = \frac{2 \ g}{20 \ g} \times 100 = 10 \%$

(B) Molarity $(M)$ is defined as the number of moles of solute per litre of solution.
$M = \frac{\text{Number of moles of solute}}{\text{Volume of solution in litres}}$
First,calculate the number of moles of $NaOH$:
$\text{Moles of } NaOH = \frac{\text{Mass}}{\text{Molar mass}} = \frac{4 \,g}{40 \,g \,mol^{-1}} = 0.1 \,mol$
Next,convert the volume of the solution to litres:
$250 \,mL = 0.250 \,L$
Now,calculate the molarity:
$M = \frac{0.1 \,mol}{0.250 \,L} = 0.4 \,mol \,L^{-1} = 0.4 \,M$

(N/A) Given: Molarity $(M)$ = $3 \ mol \ L^{-1}$,Density $(d)$ = $1.25 \ g \ mL^{-1}$,Molar mass of $NaCl$ = $58.5 \ g \ mol^{-1}$.
Mass of $NaCl$ in $1 \ L$ solution = $3 \ mol \times 58.5 \ g \ mol^{-1} = 175.5 \ g$.
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 $(m)$ = $\frac{\text{Moles of solute}}{\text{Mass of solvent in } kg} = \frac{3 \ mol}{1.0745 \ kg} \approx 2.79 \ m$.

(N/A) $(i)$ $1 \ ppm$ is equivalent to $1$ part out of $10^{6}$ parts.
Mass percent of $15 \ ppm$ chloroform in water $= \frac{15}{10^{6}} \times 100 = 1.5 \times 10^{-3} \%$.
$(ii)$ $10^{6} \ g$ of the sample contains $15 \ g$ of $CHCl_{3}$.
Mass of water $\simeq 10^{6} \ g$ (since the amount of solute is negligible).
Molar mass of $CHCl_{3} = 12.01 + 1.008 + 3(35.45) = 119.37 \ g \ mol^{-1}$.
Moles of $CHCl_{3} = \frac{15 \ g}{119.37 \ g \ mol^{-1}} \simeq 0.1256 \ mol$.
Molality $(m) = \frac{\text{moles of solute}}{\text{mass of solvent in } kg} = \frac{0.1256 \ mol}{10^{6} \ g \times 10^{-3} \ kg/g} = 1.256 \times 10^{-4} \ m$.

(N/A) $0.50 \, mol$ $Na_{2}CO_{3}$ represents a specific quantity of the substance in terms of moles. The mass of $0.50 \, mol$ $Na_{2}CO_{3}$ is calculated as: $0.50 \, mol \times 106 \, g \, mol^{-1} = 53 \, g$.
$0.50 \, M$ $Na_{2}CO_{3}$ represents the molarity of the solution,which means $0.50 \, mol$ of $Na_{2}CO_{3}$ is dissolved in $1 \, L$ of the solution.
Therefore,$0.50 \, mol$ refers to the amount of substance,while $0.50 \, M$ refers to the concentration of the solution.

(N/A) Mole fraction of $C_2H_5OH = \frac{\text{Number of moles of } C_2H_5OH}{\text{Number of moles of solution}}$
$0.040 = \frac{n_{C_2H_5OH}}{n_{C_2H_5OH} + n_{H_2O}}$ ...........$(I)$
Number of moles present in $1 \, L$ water:
$n_{H_2O} = \frac{1000 \, g}{18 \, g \, mol^{-1}} = 55.55 \, mol$
Substituting the value of $n_{H_2O}$ in equation $(I)$:
$\frac{n_{C_2H_5OH}}{n_{C_2H_5OH} + 55.55} = 0.040$
$n_{C_2H_5OH} = 0.040 \, n_{C_2H_5OH} + 2.222$
$0.96 \, n_{C_2H_5OH} = 2.222 \, mol$
$n_{C_2H_5OH} = \frac{2.222}{0.96} \, mol = 2.314 \, mol$
Since the volume of the solution is approximately $1 \, L$ (assuming dilute solution density $\approx 1 \, g/mL$):
Molarity $= \frac{2.314 \, mol}{1 \, L} = 2.314 \, M$

(A) Assume that we have $100 \ g$ of solution.
Solution will contain $20 \ g$ of ethylene glycol and $80 \ g$ of water.
Molar mass of $C_{2}H_{6}O_{2} = 12 \times 2 + 1 \times 6 + 16 \times 2 = 62 \ g \ mol^{-1}$.
Moles of $C_{2}H_{6}O_{2} = \frac{20 \ g}{62 \ g \ mol^{-1}} = 0.322 \ mol$.
Moles of water = $\frac{80 \ g}{18 \ g \ mol^{-1}} = 4.444 \ mol$.
Mole fraction of ethylene glycol $(x_{glycol})$ = $\frac{\text{moles of } C_{2}H_{6}O_{2}}{\text{moles of } C_{2}H_{6}O_{2} + \text{moles of } H_{2}O}$.
$x_{glycol} = \frac{0.322}{0.322 + 4.444} = \frac{0.322}{4.766} = 0.068$.

(B) The molar mass of $NaOH = 23 + 16 + 1 = 40 \, g \, mol^{-1}$.
Moles of $NaOH = \frac{\text{mass}}{\text{molar mass}} = \frac{5 \, g}{40 \, g \, mol^{-1}} = 0.125 \, mol$.
Volume of the solution in litres $= \frac{450 \, mL}{1000 \, mL \, L^{-1}} = 0.450 \, L$.
Molarity $(M) = \frac{\text{moles of solute}}{\text{volume of solution in litres}} = \frac{0.125 \, mol}{0.450 \, L} = 0.2777... \, M \approx 0.278 \, M$.

(A) Molar mass of $CH_{3}COOH = (2 \times 12) + (4 \times 1) + (2 \times 16) = 60 \ g \ mol^{-1}$.
Moles of $CH_{3}COOH = \frac{2.5 \ g}{60 \ g \ mol^{-1}} = 0.04167 \ mol$.
Mass of solvent (benzene) in $kg = \frac{75 \ g}{1000} = 0.075 \ kg$.
Molality $(m) = \frac{\text{Moles of solute}}{\text{Mass of solvent in } kg} = \frac{0.04167 \ mol}{0.075 \ kg} = 0.556 \ mol \ kg^{-1}$.

Mass percentage of $C_{6}H_{6} = \frac{\text{Mass of } C_{6}H_{6}}{\text{Total mass of the solution}} \times 100 \%$
$= \frac{22 \ g}{22 \ g + 122 \ g} \times 100 \%$
$= \frac{22}{144} \times 100 \% = 15.28 \%$
Mass percentage of $CCl_{4} = \frac{\text{Mass of } CCl_{4}}{\text{Total mass of the solution}} \times 100 \%$
$= \frac{122 \ g}{22 \ g + 122 \ g} \times 100 \%$
$= \frac{122}{144} \times 100 \% = 84.72 \%$
Alternatively,
Mass percentage of $CCl_{4} = (100 - 15.28) \% = 84.72 \%$

(A) Let the total mass of the solution be $100 \, g$ and the mass of benzene be $30 \, g$.
$\therefore$ Mass of carbon tetrachloride $= (100 - 30) \, g = 70 \, g$.
Molar mass of benzene $(C_6H_6) = (6 \times 12 + 6 \times 1) \, g \, mol^{-1} = 78 \, g \, mol^{-1}$.
Number of moles of benzene $= \frac{30}{78} \, mol \approx 0.3846 \, mol$.
Molar mass of $CCl_4 = (12 + 4 \times 35.5) \, g \, mol^{-1} = 154 \, g \, mol^{-1}$.
$\therefore$ Number of moles of $CCl_4 = \frac{70}{154} \, mol \approx 0.4545 \, mol$.
Thus,the mole fraction of $C_6H_6$ is given as:
$x_{C_6H_6} = \frac{n_{C_6H_6}}{n_{C_6H_6} + n_{CCl_4}} = \frac{0.3846}{0.3846 + 0.4545} = \frac{0.3846}{0.8391} \approx 0.458$.

(D) Molar mass of urea $(NH_2CONH_2) = 2(14 + 2 \times 1) + 12 + 16 + 14 + 2 = 60 \, g \, mol^{-1}$.
$0.25 \, \text{molal}$ aqueous solution means $0.25 \, mol$ of urea is dissolved in $1000 \, g$ of water.
Mass of urea $= 0.25 \, mol \times 60 \, g \, mol^{-1} = 15 \, g$.
Total mass of solution $= \text{Mass of solute} + \text{Mass of solvent} = 15 \, g + 1000 \, g = 1015 \, g$.
In $1015 \, g$ of solution,urea present is $15 \, g$.
Therefore,in $2500 \, g$ $(2.5 \, kg)$ of solution,mass of urea $= \frac{15 \times 2500}{1015} \, g \approx 36.95 \, g \approx 37 \, g$.

(N/A) Molar mass of $KI = 39 + 127 = 166 \, g \, mol^{-1}$.
$20 \%$ (mass/mass) aqueous solution of $KI$ means $20 \, g$ of $KI$ is present in $100 \, g$ of solution. Thus,$20 \, g$ of $KI$ is present in $(100 - 20) \, g = 80 \, g$ of water.
Molality $(m) = \frac{\text{Moles of } KI}{\text{Mass of water in } kg} = \frac{20 / 166}{0.080} \, m = 1.506 \, m \approx 1.51 \, m$.
$(b)$ Density of solution $= 1.202 \, g \, mL^{-1}$.
Volume of $100 \, g$ solution $= \frac{\text{Mass}}{\text{Density}} = \frac{100}{1.202} \, mL = 83.19 \, mL = 0.08319 \, L$.
Molarity $(M) = \frac{\text{Moles of } KI}{\text{Volume of solution in } L} = \frac{20 / 166}{0.08319} \, M = 1.448 \, M \approx 1.45 \, M$.
$(c)$ Moles of $KI = \frac{20}{166} = 0.1205 \, mol$.
Moles of water $= \frac{80}{18} = 4.444 \, mol$.
Mole fraction of $KI = \frac{n_{KI}}{n_{KI} + n_{H_2O}} = \frac{0.1205}{0.1205 + 4.444} = \frac{0.1205}{4.5645} = 0.0264$.

(N/A) solution is a homogeneous mixture of two or more chemically non-reacting substances.
There are $9$ types of solutions based on the physical state of the solute and solvent:
$1$. Gaseous solutions:
$(a)$ Gas in gas: Air (mixture of $O_2$ and $N_2$).
$(b)$ Liquid in gas: Water vapour in air.
$(c)$ Solid in gas: Camphor vapours in $N_2$ gas.
$2$. Liquid solutions:
$(a)$ Gas in liquid: $CO_2$ dissolved in water (aerated water).
$(b)$ Liquid in liquid: Ethanol dissolved in water.
$(c)$ Solid in liquid: Sugar dissolved in water.
$3$. Solid solutions:
$(a)$ Gas in solid: Hydrogen gas adsorbed on Palladium $(Pd)$.
$(b)$ Liquid in solid: Amalgam of Mercury with Sodium $(Na-Hg)$.
$(c)$ Solid in solid: Gold ornaments (Copper or Silver alloyed with Gold).

(N/A) $(i)$ Mole Fraction: It is defined as the ratio of the number of moles of the solute to the total number of moles in the solution. If $n_A$ is the number of moles of solute and $n_B$ is the number of moles of solvent,then the mole fraction of solute $(X_A)$ is given by:
$X_A = \frac{n_A}{n_A + n_B}$
$(ii)$ Molality $(m)$: It is defined as the number of moles of solute present in $1 \ kg$ $(1000 \ g)$ of solvent.
$m = \frac{\text{Number of moles of solute}}{\text{Mass of solvent in kg}}$
$(iii)$ Molarity $(M)$: It is defined as the number of moles of solute present in $1 \ L$ of solution.
$M = \frac{\text{Number of moles of solute}}{\text{Volume of solution in litres}}$
$(iv)$ Mass Percentage: It is defined as the mass of solute in grams present in $100 \ g$ of solution.
$\text{Mass percentage} = \frac{\text{Mass of solute}}{\text{Mass of solution}} \times 100$

(N/A) $10 \%$ $w/w$ solution of glucose in water means $10 \ g$ of glucose and $90 \ g$ of water.
$10 \ g$ of glucose $= \frac{10}{180} = 0.0555 \ mol$
$90 \ g$ of $H_2O = \frac{90}{18} = 5 \ mol$
$\text{Molality} = \frac{\text{Moles of solute} \times 1000}{\text{Mass of solvent in grams}} = \frac{0.0555}{90} \times 1000 = 0.617 \ m$
$\text{Mole fraction of glucose } (X_g) = \frac{0.0555}{5 + 0.0555} = 0.01$
$\text{Mole fraction of water } (X_w) = \frac{5}{5 + 0.0555} = 0.99$
$\text{Volume of } 100 \ g \text{ solution} = \frac{100}{1.2} = 83.33 \ mL$
$\text{Molarity} = \frac{0.0555}{83.33} \times 1000 = 0.67 \ M$

(A) Total amount of solute present in the mixture is given by,
$300 \times \frac{25}{100} + 400 \times \frac{40}{100} = 75 + 160 = 235 \, g$
Total amount of solution $= 300 + 400 = 700 \, g$
Therefore,mass percentage $(w/w)$ of the solute in the resulting solution $= \frac{235}{700} \times 100 \% = 33.57 \%$

(N/A) Mass of solute $= 222.6 \ g$
Molar mass of solute $C_2H_6O_2 = (12 \times 2) + (1 \times 6) + (16 \times 2) = 62 \ g \ mol^{-1}$
Moles of solute $= \frac{222.6 \ g}{62 \ g \ mol^{-1}} = 3.59 \ mol$
Mass of solvent $= 200 \ g = 0.2 \ kg$
Molality $(m) = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{3.59 \ mol}{0.2 \ kg} = 17.95 \ mol \ kg^{-1}$
Total mass of solution $= 222.6 \ g + 200 \ g = 422.6 \ g$
Volume of solution $= \frac{\text{mass of solution}}{\text{density}} = \frac{422.6 \ g}{1.072 \ g \ mL^{-1}} = 394.21 \ mL = 0.39421 \ L$
Molarity $(M) = \frac{\text{moles of solute}}{\text{volume of solution in L}} = \frac{3.59 \ mol}{0.39421 \ L} = 9.11 \ mol \ L^{-1}$

(N/A) $15$ $ppm$ means $15$ parts in $10^{6}$ parts by mass in the solution.
$(i)$ Percentage by mass $= \frac{15}{10^{6}} \times 100 = 1.5 \times 10^{-3} \%$.
$(ii)$ As $15 \, g$ of chloroform is present in $10^{6} \, g$ of the solution,the mass of the solvent is approximately $10^{6} \, g$ (since $15 \, g$ is negligible compared to $10^{6} \, g$).
Molar mass of $CHCl_{3} = 12 + 1 + 3 \times 35.5 = 119.5 \, g \, mol^{-1}$.
Moles of $CHCl_{3} = \frac{15 \, g}{119.5 \, g \, mol^{-1}} \approx 0.1255 \, mol$.
Molality $(m) = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.1255 \, mol}{10^{6} \, g / 1000} = \frac{0.1255}{1000} \, mol \, kg^{-1} = 1.255 \times 10^{-4} \, m$.

(B) The molality $(m)$ is defined as the number of moles of solute per kilogram of solvent.
Given mass of $Na^{+}$ ions $= 92\, g$.
Molar mass of $Na^{+}$ ions $= 23\, g\, mol^{-1}$.
Number of moles of $Na^{+}$ ions $= \frac{92\, g}{23\, g\, mol^{-1}} = 4\, mol$.
Mass of solvent (water) $= 1\, kg$.
Molality $(m)$ $= \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{4\, mol}{1\, kg} = 4\, m$.

(N/A) The mass of the solute (aspirin,$C_{9}H_{8}O_{4}$) is $6.5 \ g$.
The mass of the solvent (acetonitrile,$CH_{3}CN$) is $450 \ g$.
The total mass of the solution is the sum of the mass of the solute and the solvent: $6.5 \ g + 450 \ g = 456.5 \ g$.
The mass percentage of $C_{9}H_{8}O_{4}$ is calculated as: $\frac{\text{mass of solute}}{\text{total mass of solution}} \times 100$.
Mass percentage $= \frac{6.5 \ g}{456.5 \ g} \times 100 \% \approx 1.424 \%$.

(N/A) The molar mass of nalorphene $(C_{19}H_{21}NO_{3})$ is calculated as:
$19 \times 12 + 21 \times 1 + 14 + 3 \times 16 = 311 \, g \, mol^{-1}$
In a $1.5 \times 10^{-3} \, m$ aqueous solution,$1000 \, g$ of water contains $1.5 \times 10^{-3} \, mol$ of nalorphene.
Mass of nalorphene $= 1.5 \times 10^{-3} \, mol \times 311 \, g \, mol^{-1} = 0.4665 \, g$.
Total mass of the solution $= \text{Mass of solvent} + \text{Mass of solute} = 1000 \, g + 0.4665 \, g = 1000.4665 \, g$.
This means $1000.4665 \, g$ of solution contains $0.4665 \, g$ of nalorphene.
To find the mass of solution containing $1.5 \, mg$ $(1.5 \times 10^{-3} \, g)$ of nalorphene:
$\text{Mass of solution} = \frac{1000.4665 \, g \times 1.5 \times 10^{-3} \, g}{0.4665 \, g} \approx 3.22 \, g$.
Thus,the mass of the aqueous solution required is $3.22 \, g$.

(N/A) $0.15 \ M$ solution of benzoic acid in methanol means that $1000 \ mL$ of solution contains $0.15 \ mol$ of benzoic acid.
Therefore,$250 \ mL$ of solution contains $= \frac{0.15 \times 250}{1000} \ mol = 0.0375 \ mol$ of benzoic acid.
Molar mass of benzoic acid $(C_{6}H_{5}COOH) = (7 \times 12) + (6 \times 1) + (2 \times 16) = 122 \ g \ mol^{-1}$.
Hence,the required mass of benzoic acid $= 0.0375 \ mol \times 122 \ g \ mol^{-1} = 4.575 \ g$.

(N/A) Solutions are classified into $9$ types based on the physical states of solute and solvent:
$1$. Gas in Gas: Example,mixture of $O_2$ and $N_2$ gases.
$2$. Liquid in Gas: Example,chloroform mixed with $N_2$ gas.
$3$. Solid in Gas: Example,camphor in $N_2$ gas.
$4$. Gas in Liquid: Example,$O_2$ dissolved in water.
$5$. Liquid in Liquid: Example,ethanol dissolved in water.
$6$. Solid in Liquid: Example,glucose dissolved in water.
$7$. Gas in Solid: Example,solution of $H_2$ in palladium.
$8$. Liquid in Solid: Example,amalgam of mercury with sodium.
$9$. Solid in Solid: Example,copper dissolved in gold (alloys).

(C) The physical state of a solution is determined by the physical state of the solvent.
Since $N_2$ gas is the solvent,the resulting solution of camphor $(solid)$ in $N_2$ gas $(gas)$ exists in the $gaseous$ state.

(N/A) binary solution is a homogeneous mixture consisting of exactly two components: one solute and one solvent.
$1$. Solute: The component present in a smaller quantity.
$2$. Solvent: The component present in a larger quantity.
For example,a sugar solution is a binary solution where sugar is the solute and water is the solvent.

(N/A) The concentration of a solution is defined as the amount of solute present in a given quantity (either volume or mass) of the solution or solvent.
Different methods to express concentration include:
$1$. Molarity $(M)$: Moles of solute per liter of solution.
$2$. Molality $(m)$: Moles of solute per kilogram of solvent.
$3$. Normality $(N)$: Gram equivalent of solute per liter of solution.
$4$. Mole fraction $(x)$: Ratio of moles of one component to the total moles of all components.
$5$. Mass percentage $(\% w/w)$: Mass of solute in $100 \ g$ of solution.
$6$. Volume percentage $(\% v/v)$: Volume of solute in $100 \ mL$ of solution.
$7$. Parts per million $(ppm)$: Parts of solute per million parts of solution.

(N/A) The volume of solute $(mL)$ dissolved in $100 \ mL$ of solution is expressed as volume percentage $(\% \ V/V)$.
$\text{Volume } \% \text{ of a component} = \frac{\text{Volume of the component} \times 100}{\text{Total volume of solution}}$
For example,a $10 \% \text{ ethanol solution in water means that } 10 \ mL \text{ of ethanol is dissolved in water such that the total volume of the solution is } 100 \ mL$.
Solutions containing liquids are commonly expressed in this unit.
For example,a $35 \% \ (V/V)$ solution of ethylene glycol,an antifreeze,is used in cars for cooling the engine.
At this concentration,the antifreeze lowers the freezing point of water to $255.4 \ K$ $(-17.6^{\circ} C)$.

(N/A) Mass by volume percentage $(\% \, w/V)$: It is defined as the mass of the solute in grams dissolved in $100 \, mL$ of the solution.
Mass percentage $(\% \, w/w)$: It is defined as the mass of the component in grams per $100 \, g$ of the solution.
Formula: $\text{Mass } \% \text{ of a component} = \frac{\text{Mass of the component in the solution}}{\text{Total mass of the solution}} \times 100$.
Example for $(\% \, w/w)$: If a solution is $10 \% \, w/w$ glucose in water,it means $10 \, g$ of glucose is dissolved in $90 \, g$ of water,making a total of $100 \, g$ of solution.

(N/A) When a solute is present in trace quantities,it is convenient to express concentration in parts per million $(ppm)$.
Parts per million = $\frac{\text{Number of parts of the component}}{\text{Total number of parts of all components}} \times 10^{6}$
As in the case of percentage,concentration in $ppm$ can also be expressed as mass to mass,volume to volume,or mass to volume.
For example,a litre of sea water (which weighs $1030 \ g$) contains about $6 \times 10^{-3} \ g$ of dissolved oxygen $(O_2)$. Such a small concentration is expressed as $5.8 \ g$ per $10^{6} \ g$ $(5.8 \ ppm)$ of seawater. The concentration of pollutants in water or atmosphere is often expressed in terms of $\mu g \ mL^{-1}$ or $ppm$.

(N/A) Definition: The ratio of the number of moles of a specific component to the total number of moles of all components in a solution is known as the mole fraction of that component.
The symbol used for mole fraction is $x$,and the subscript denotes the specific component.
Mole fraction of a component $= \frac{\text{Number of moles of the component}}{\text{Total number of moles of all the components}}$
For a binary mixture with $n_{A}$ moles of component $A$ and $n_{B}$ moles of component $B$,the mole fraction of $A$ is given by:
$x_{A} = \frac{n_{A}}{n_{A} + n_{B}}$
For a solution containing $i$ components,the mole fraction of component $i$ is:
$x_{i} = \frac{n_{i}}{\sum n_{i}}$
The sum of all mole fractions in a solution is always unity:
$x_{1} + x_{2} + \ldots + x_{i} = 1$
Mole fraction is useful in relating physical properties like vapour pressure to concentration and is essential for calculations involving gas mixtures.

(N/A) Molarity $(M)$ is defined as the number of moles of solute dissolved in one litre (or one cubic decimeter) of solution.
Molarity $= \frac{\text{Moles of solute}}{\text{Volume of solution in litre}}$
For example,a $0.25 \ mol \ L^{-1}$ (or $0.25 \ M$) solution of $NaOH$ means that $0.25 \ mol$ of $NaOH$ is dissolved in one litre of solution.
Molality $(m)$ is defined as the number of moles of solute per kilogram $(kg)$ of solvent.
Molality $(m)$ $= \frac{\text{Moles of solute}}{\text{Mass of solvent in } kg}$
For example,a $1.00 \ mol \ kg^{-1}$ (or $1.00 \ m$) solution of $KCl$ means that $1 \ mol$ $(74.5 \ g)$ of $KCl$ is dissolved in $1 \ kg$ of water.
Note: Mass percentage,ppm,mole fraction,and molality are independent of temperature,whereas molarity depends on temperature because volume changes with temperature.

(A) $1$. Calculate the moles of $NaCl$: $n_{NaCl} = \frac{\text{mass}}{\text{molar mass}} = \frac{5.85 \ g}{58.5 \ g/mol} = 0.1 \ mol$.
$2$. Calculate the moles of water $(H_2O)$: $n_{H_2O} = \frac{90 \ g}{18 \ g/mol} = 5 \ mol$.
$3$. Calculate the mole fraction of $NaCl$ $(x_{NaCl})$: $x_{NaCl} = \frac{n_{NaCl}}{n_{NaCl} + n_{H_2O}} = \frac{0.1}{0.1 + 5} = \frac{0.1}{5.1} \approx 0.0196$.

(A) $5.2 \ m$ solution means $5.2 \ mol$ of methanol is dissolved in $1 \ kg$ $(1000 \ g)$ of water.
Number of moles of methanol $(n_{methanol})$ = $5.2 \ mol$.
Number of moles of water $(n_{water})$ = $\frac{1000 \ g}{18 \ g/mol} \approx 55.56 \ mol$.
Mole fraction of methanol $(x_{methanol})$ = $\frac{n_{methanol}}{n_{methanol} + n_{water}} = \frac{5.2}{5.2 + 55.56} = \frac{5.2}{60.76} \approx 0.0856$.
Rounding to three decimal places,we get $0.086$.

(A) $1$. Calculate the molar mass of glycerine $(C_3H_8O_3)$: $(3 \times 12) + (8 \times 1) + (3 \times 16) = 36 + 8 + 48 = 92 \ g/mol$.
$2$. Calculate the moles of glycerine: $n_{\text{glycerine}} = \frac{46 \ g}{92 \ g/mol} = 0.5 \ mol$.
$3$. Calculate the moles of water $(H_2O)$: $n_{\text{water}} = \frac{36 \ g}{18 \ g/mol} = 2.0 \ mol$.
$4$. Calculate the total moles in the solution: $n_{\text{total}} = 0.5 + 2.0 = 2.5 \ mol$.
$5$. Calculate the mole fraction of glycerine $(X_{\text{glycerine}})$: $X_{\text{glycerine}} = \frac{n_{\text{glycerine}}}{n_{\text{total}}} = \frac{0.5}{2.5} = 0.2$.

(B) Molarity $(M)$ is defined as the number of moles of solute per liter of solution.
Given:
Moles of solute $(n)$ = $0.01 \ mol$
Mass of solvent (water) = $10 \ g$
Assuming the density of the solution is approximately $1 \ g/mL$,the volume of the solution is $10 \ mL = 0.01 \ L$.
Molarity $(M)$ = $\frac{n}{V(L)} = \frac{0.01 \ mol}{0.01 \ L} = 1 \ M$.

(N/A) The expression $10\%$ glucose by weight means that $10 \ g$ of glucose is dissolved in $90 \ g$ of water,resulting in a total mass of $100 \ g$ of the solution.
Mathematically,it is defined as:
$\text{Mass percentage} = \frac{\text{Mass of solute}}{\text{Mass of solution}} \times 100$
$\frac{10 \ g \text{ glucose}}{100 \ g \text{ solution}} \times 100 = 10\%$.

(N/A) Percentage by weight = $\frac{\text{Mass of solute}}{\text{Mass of solution}} \times 100$