Method of expressing concentration of solution Questions in English

(A) The formula for molality $(m)$ is given by: $m = \frac{w_2}{M_2} \times \frac{1000}{w_1(g)}$
Where $w_2 = 15.20 \ g$ (mass of solute),$M_2 = 60 \ g \ mol^{-1}$ (molar mass of urea),and $w_1 = 150 \ g$ (mass of solvent).
Substituting the values:
$m = \frac{15.20}{60} \times \frac{1000}{150}$
$m = 0.2533 \times 6.6667$
$m = 1.689 \ mol \ kg^{-1}$

(A) $1.25$ molal solution means $1.25$ moles of $NaOH$ are present in $1000 \ g$ of solvent.
Molar mass of $NaOH = 40 \ g/mol$.
Mass of $NaOH = 1.25 \times 40 = 50 \ g$.
Mass of solvent $= 1000 \ g$.
Total mass of solution $= 1000 + 50 = 1050 \ g$.
Percentage by weight $= \frac{50}{1050} \times 100 = 4.76 \%$.

(A) Gasoline is a mixture of liquid hydrocarbons.
In a liquid-liquid solution,the solute is a liquid and the solvent is also a liquid.
Since gasoline consists of various liquid hydrocarbons mixed together,it is classified as a liquid-in-liquid solution.
Therefore,the correct option is $A$.

(C) Sea water is a solution where salt (a solid solute) is dissolved in water (a liquid solvent). Therefore,it is classified as a $solid \text{ in } liquid$ type of solution.

(D) Air is a homogeneous mixture of various gases like $N_2$,$O_2$,$Ar$,$CO_2$,etc.
Since both the solute and the solvent are in the gaseous state,it is classified as a solution of $gas$ in $gas$.

(A) In a solution of chloroform in nitrogen,chloroform $(CHCl_3)$ is the solute (liquid state) and nitrogen $(N_2)$ is the solvent (gaseous state).
Therefore,it is an example of a liquid in gas type of solution.

(B) The molar mass of Glucose $(C_6H_{12}O_6)$ is $(6 \times 12) + (12 \times 1) + (6 \times 16) = 72 + 12 + 96 = 180 \ g/mol$.
Molality $(m)$ is defined as the number of moles of solute per kilogram of solvent.
Given $m = 0.25 \ mol/kg$.
Let the mass of the solvent (water) be $W \ kg$.
The mass of the solution is $W + \text{mass of glucose} = 2.5 \ kg$.
Mass of glucose $= \text{moles} \times \text{molar mass} = (0.25 \times W) \times 180 = 45W \ g = 0.045W \ kg$.
Substituting this into the solution mass equation: $W + 0.045W = 2.5 \ kg$.
$1.045W = 2.5 \implies W = 2.5 / 1.045 \approx 2.3923 \ kg$.
Mass of glucose $= 0.045 \times 2.3923 \ kg \approx 0.10765 \ kg = 107.65 \ g$.

(B) $1 \ m$ solution means $1 \ \text{mole}$ of solute (urea) is dissolved in $1 \ \text{kg}$ $(1000 \ \text{g})$ of solvent (water).
Number of moles of urea $(n_{\text{urea}})$ = $1 \ \text{mol}$.
Number of moles of water $(n_{\text{water}})$ = $\frac{1000 \ \text{g}}{18 \ \text{g/mol}} \approx 55.56 \ \text{mol}$.
Mole fraction of urea $(x_{\text{urea}})$ = $\frac{n_{\text{urea}}}{n_{\text{urea}} + n_{\text{water}}}$.
$x_{\text{urea}} = \frac{1}{1 + 55.56} = \frac{1}{56.56} \approx 0.01768$.

(B) The molar mass of ethanoic acid $(CH_3COOH)$ is $(2 \times 12) + (4 \times 1) + (2 \times 16) = 24 + 4 + 32 = 60 \ g/mol$.
Number of moles of ethanoic acid = $\frac{\text{mass}}{\text{molar mass}} = \frac{3.0 \ g}{60 \ g/mol} = 0.05 \ mol$.
Mass of solvent (benzene) = $50 \ g = 0.05 \ kg$.
Molality $(m)$ = $\frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.05 \ mol}{0.05 \ kg} = 1.0 \ mol/kg$.
Therefore,the correct option is $B$.

(C) Given,mole fraction of solute $(x_B)$ = $0.25$.
Since it is an aqueous solution,the mole fraction of solvent (water,$x_A$) = $1 - 0.25 = 0.75$.
Molality $(m)$ is defined as the number of moles of solute per kilogram of solvent.
$m = \frac{n_B}{W_A \text{ (in kg)}} = \frac{n_B}{n_A \times M_A \times 10^{-3}}$,where $M_A$ is the molar mass of water $(18 \ g/mol)$.
$m = \frac{x_B}{x_A \times M_A \text{ (in kg/mol)}} = \frac{0.25}{0.75 \times 18 \times 10^{-3}}$.
$m = \frac{0.25}{0.0135} \approx 18.52 \ m$.

(B) Concentration units that involve volume (such as Molarity,Normality,and Formality) 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,the value of Molality remains constant regardless of temperature changes.
Therefore,the correct option is $B$.

(C) Mass of solute in the first solution $= 400 \ g \times 0.25 = 100 \ g$.
Mass of solute in the second solution $= 600 \ g \times 0.40 = 240 \ g$.
Total mass of solute $= 100 \ g + 240 \ g = 340 \ g$.
Total mass of the resulting solution $= 400 \ g + 600 \ g = 1000 \ g$.
Mass percentage of the resulting solution $= (\text{Total mass of solute} / \text{Total mass of solution}) \times 100 = (340 \ g / 1000 \ g) \times 100 = 34 \%$.
Therefore,the correct option is $C$.

(C) The mass percentage of a solute is calculated using the formula: $\text{Mass percentage} = \frac{\text{Mass of solute}}{\text{Mass of solute} + \text{Mass of solvent}} \times 100$.
Given: $\text{Mass of benzene (solute)} = 22 \ g$.
$\text{Mass of carbon tetrachloride (solvent)} = 122 \ g$.
$\text{Total mass of solution} = 22 \ g + 122 \ g = 144 \ g$.
$\text{Mass percentage of benzene} = \frac{22}{144} \times 100 = 15.277 \% \approx 15.28 \%$.
Therefore,the correct option is $C$.

(B) Given: $10\%$ w/w aqueous solution of $NaOH$ means $10 \ g$ of $NaOH$ is present in $100 \ g$ of solution.
Mass of solute $(NaOH)$ = $10 \ g$.
Mass of solvent $(H_2O)$ = $100 \ g - 10 \ g = 90 \ g = 0.09 \ kg$.
Molar mass of $NaOH$ = $40 \ g \ mol^{-1}$.
Number of moles of $NaOH$ = $\frac{10 \ g}{40 \ g \ mol^{-1}} = 0.25 \ mol$.
Molality $(m)$ = $\frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.25 \ mol}{0.09 \ kg} \approx 2.78 \ m$.

(C) To find the molarity of the diluted solution,we use the dilution formula: $M_1 \cdot V_1 = M_2 \cdot V_2$
Given:
$M_1 = 0.5 \ M$
$V_1 = 30 \ mL$
$V_2 = 500 \ mL$
Substituting the values into the formula:
$(0.5 \ M) \cdot (30 \ mL) = M_2 \cdot (500 \ mL)$
$M_2 = \frac{0.5 \times 30}{500}$
$M_2 = \frac{15}{500} = 0.03 \ M$
Therefore,the molarity of the diluted solution is $0.03 \ M$.

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

(C) Given: $30 \% \ w/w$ $NaOH$ solution.
This means $30 \ g$ of $NaOH$ is present in $100 \ g$ of the solution.
Mass of solute $(NaOH)$ $= 30 \ g$.
Mass of solvent (water) $= 100 \ g - 30 \ g = 70 \ g = 0.07 \ kg$.
Molar mass of $NaOH = 23 + 16 + 1 = 40 \ g/mol$.
Moles of $NaOH = \frac{30 \ g}{40 \ g/mol} = 0.75 \ mol$.
Molality $(m)$ $= \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.75 \ mol}{0.07 \ kg} = 10.71 \ m$.

(A) The concentration is given as $0.05 \% w/v$,which means $0.05 \ g$ of $CaCl_2$ is present in $100 \ mL$ of the solution.
To convert $w/v$ percentage to $ppm$ (parts per million),we use the formula: $ppm = (w/v \%) \times 10^4$.
$ppm = 0.05 \times 10^4 = 500 \ ppm$.
Therefore,the correct option is $A$.

(C) Concentration units that involve volume (such as $Molarity$,$Normality$,and $Formality$) 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,$Molality$ remains independent of temperature.

(C) mixture of camphor in air is a type of solution where the solute is a solid $(camphor)$ and the solvent is a gas $(air)$.
Since the physical state of the solvent determines the state of the solution, a mixture of a solid in a gas is classified as a $gas solution$.

(A) Mole fraction is the ratio of moles of one component to the total moles of all components,so it has no units.
Mass percent $(W/W)$ is the ratio of the mass of a solute to the total mass of the solution multiplied by $100$,which is also a dimensionless quantity.
Therefore,both Mole fraction and Mass percent $(W/W)$ are unitless.

(A) Let $V \ L$ of $10 \ N$ $HCl$ be mixed with $(1-V) \ L$ of $4 \ N$ $HCl$ to obtain $1 \ L$ of $7 \ N$ $HCl$.
Using the mixture formula $N_1 V_1 + N_2 V_2 = N_3 V_3$:
$10V + 4(1-V) = 7 \times 1$
$10V + 4 - 4V = 7$
$6V = 3$
$V = 0.50 \ L$
Thus,the volume of $10 \ N$ $HCl$ is $0.50 \ L$ and the volume of $4 \ N$ $HCl$ is $1 - 0.50 = 0.50 \ L$.

(D) Given,amount of ethyl alcohol $= 414 \ g$.
Amount of water $= 18 \ g$.
Molar mass of water $(H_2O) = 18 \ g/mol$.
Molar mass of ethyl alcohol $(C_2H_5OH) = 46 \ g/mol$.
Moles of $C_2H_5OH = \frac{414}{46} = 9 \ mol$.
Moles of $H_2O = \frac{18}{18} = 1 \ mol$.
Mole fraction of $H_2O = \frac{\text{Moles of } H_2O}{\text{Moles of } H_2O + \text{Moles of } C_2H_5OH} = \frac{1}{1+9} = \frac{1}{10} = 0.1$.

(B) The number of moles of solute is calculated using the formula: $n = M \times V$ (where $M$ is molarity and $V$ is volume in liters).
For $1000 \ mL$ of $1 \ M \ NaOH$,$n = 1 \ M \times 1 \ L = 1 \ mol$.
Checking the options:
$(1)$ $0.2 \ M \times 0.2 \ L = 0.04 \ mol$.
$(2)$ $1 \ M \times 0.2 \ L = 0.2 \ mol$ and $0.2 \ M \times 1 \ L = 0.2 \ mol$. Since $0.2 = 0.2$,this statement is true.
$(3)$ $0.2 \ M \times 0.1 \ L = 0.02 \ mol$.
$(4)$ $0.2 \ M \times 2 \ L = 0.4 \ mol$.

(B) Mass percent of solute $= \frac{\text{Mass of solute}}{\text{Mass of solution}} \times 100$
$= \frac{3 \ g}{3 \ g + 18 \ g} \times 100$
$= \frac{3}{21} \times 100$
$= \frac{1}{7} \times 100 \approx 14.28 \%$

(C) Step $1$: Calculate the molar mass of ethylene glycol $(C_2H_6O_2)$.
$M = (2 \times 12) + (6 \times 1) + (2 \times 16) = 24 + 6 + 32 = 62 \ g \ mol^{-1}$.
Step $2$: Calculate the number of moles of ethylene glycol.
$n = \frac{\text{mass}}{\text{molar mass}} = \frac{248 \ g}{62 \ g \ mol^{-1}} = 4 \ mol$.
Step $3$: Calculate the molality $(m)$ of the solution.
$m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{4 \ mol}{0.2 \ kg} = 20 \ m$.

(C) The mole fraction of a component is given by $x_i = \frac{n_i}{n_{total}}$.
Let the total number of moles be $n_{total} = 1 \ mol$.
Then,the moles of glucose $(n_{glucose})$ = $0.0244 \ mol$ and the moles of water $(n_{water})$ = $0.9756 \ mol$.
The molar mass of glucose $(C_6H_{12}O_6)$ is $180 \ g/mol$ and the molar mass of water $(H_2O)$ is $18 \ g/mol$.
Mass of glucose = $0.0244 \ mol \times 180 \ g/mol = 4.392 \ g$.
Mass of water = $0.9756 \ mol \times 18 \ g/mol = 17.5608 \ g$.
Total mass of the solution = $4.392 \ g + 17.5608 \ g = 21.9528 \ g$.
Weight percentage $(w/w)$ of glucose = $\frac{\text{mass of glucose}}{\text{total mass of solution}} \times 100 = \frac{4.392}{21.9528} \times 100 \approx 20\%$.

(C) Let $n_g$ be the moles of glucose and $n_w$ be the moles of water. Given that the mole fraction of water $(x_w)$ is $40$ times the mole fraction of glucose $(x_g)$,we have $x_w = 40 x_g$. Since $x_w + x_g = 1$,we get $40 x_g + x_g = 1$,which implies $41 x_g = 1$,so $x_g = 1/41$ and $x_w = 40/41$. The ratio of moles is $n_w / n_g = x_w / x_g = 40/1$. Thus,$n_w = 40 n_g$. The mass of glucose $(m_g)$ is $n_g \times 180 \ g/mol$ and the mass of water $(m_w)$ is $n_w \times 18 \ g/mol = 40 n_g \times 18 \ g/mol = 720 n_g \ g$. The weight percentage of glucose is $(m_g / (m_g + m_w)) \times 100 = (180 n_g / (180 n_g + 720 n_g)) \times 100 = (180 / 900) \times 100 = (1/5) \times 100 = 20\%$. Therefore,the correct option is $C$.

(B) Given,$10 \% (w/w)$ aqueous glucose solution means $10 \ g$ of glucose is present in $100 \ g$ of the solution.
Mass of solute (glucose) $= 10 \ g$.
Mass of solvent (water) $= 100 \ g - 10 \ g = 90 \ g = 0.09 \ kg$.
Molar mass of glucose $= 180 \ g \ mol^{-1}$.
Number of moles of glucose $= \frac{10 \ g}{180 \ g \ mol^{-1}} = 0.0556 \ mol$.
Molality $(m) = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.0556 \ mol}{0.09 \ kg} \approx 0.617 \ m$.
Rounding to the nearest option,the value is $0.62 \ m$.

(D) The molality $(m)$ of a solution is defined as the number of moles of solute per kilogram of solvent.
Given:
Mass of ethylene glycol $= 124 \ g$.
Molar mass of ethylene glycol $= 62 \ g \ mol^{-1}$.
Moles of ethylene glycol $(n)$ $= \frac{124 \ g}{62 \ g \ mol^{-1}} = 2 \ mol$.
Molality $(m)$ $= 10 \ mol \ kg^{-1}$.
Formula: $m = \frac{n_{\text{solute}}}{W_{\text{solvent (in kg)}}}$.
Substituting the values: $10 = \frac{2}{W_{\text{solvent (in kg)}}}$.
$W_{\text{solvent (in kg)}} = \frac{2}{10} = 0.2 \ kg$.
Since $1 \ kg = 1000 \ g$,$W_{\text{solvent}} = 0.2 \times 1000 = 200 \ g$.
Therefore,$x = 200$.

(D) $ppm$ (parts per million) $= \frac{\text{mass of solute in mg}}{\text{volume of solution in L}}$
Given,
$ppm = 1000$
Volume $= 1 \ L$
Mass of solute $= 1000 \ mg = 1 \ g$
Molar mass of $CaCO_3 = 40 + 12 + (3 \times 16) = 100 \ g \ mol^{-1}$
Number of moles $= \frac{\text{mass}}{\text{molar mass}} = \frac{1 \ g}{100 \ g \ mol^{-1}} = 0.01 \ mol = 10^{-2} \ mol$
Concentration $= \frac{\text{moles}}{\text{volume in L}} = \frac{10^{-2} \ mol}{1 \ L} = 10^{-2} \ mol \ L^{-1}$

(B) Mass of urea in the first solution $= 200 \ g \times \frac{20}{100} = 40 \ g$.
Mass of urea in the second solution $= 400 \ g \times \frac{40}{100} = 160 \ g$.
Total mass of urea $= 40 \ g + 160 \ g = 200 \ g$.
Total mass of the resultant solution $= 200 \ g + 400 \ g = 600 \ g$.
Weight percentage $(w/w \%)$ of the resultant solution $= \frac{\text{Total mass of urea}}{\text{Total mass of solution}} \times 100 = \frac{200}{600} \times 100 = 33.33 \% $.

(D) Given: Density $(\delta)$ $= 1.5 \ g \ mL^{-1}$.
Weight percentage of $HNO_3 = 68 \%$.
Assume $1 \ L$ $(1000 \ mL)$ of solution.
Mass of solution $= \text{density} \times \text{volume} = 1.5 \ g \ mL^{-1} \times 1000 \ mL = 1500 \ g$.
Mass of $HNO_3 = 68 \% \text{ of } 1500 \ g = 0.68 \times 1500 \ g = 1020 \ g$.
Molar mass of $HNO_3 = 1 + 14 + (3 \times 16) = 63 \ g \ mol^{-1}$.
Moles of $HNO_3 = \frac{1020 \ g}{63 \ g \ mol^{-1}} \approx 16.19 \ mol$.
Molarity $(M)$ $= \frac{\text{moles of solute}}{\text{volume of solution in } L} = \frac{16.19 \ mol}{1 \ L} = 16.19 \ M \approx 16.2 \ M$.

(B) Mass of urea $= 6\% \text{ of } 1000 \ g = \frac{6}{100} \times 1000 = 60 \ g$.
$Molar \text{ mass of urea } (NH_2CONH_2) = 14 + 2 + 12 + 16 + 14 + 2 = 60 \ g \ mol^{-1}$.
Moles of urea $= \frac{60 \ g}{60 \ g \ mol^{-1}} = 1 \ mol$.
Since the density of the solution is assumed to be $1.0 \ g \ mL^{-1}$,the volume of $1000 \ g$ solution $= 1000 \ mL = 1 \ L$.
Molarity $= \frac{\text{Moles of solute}}{\text{Volume of solution in } L} = \frac{1 \ mol}{1 \ L} = 1.0 \ mol \ L^{-1}$.

(C) The concentration of $NO_3^{-}$ ions is determined by the total milliequivalents of $NO_3^{-}$ divided by the total volume of the solution in $mL$.
For $Ca(NO_3)_2$: Molarity $= 0.1 \ M$. Since $1 \ mol$ of $Ca(NO_3)_2$ gives $2 \ mol$ of $NO_3^{-}$,the molarity of $NO_3^{-}$ is $0.2 \ M$. Milliequivalents of $NO_3^{-} = 0.2 \ M \times 100 \ mL = 20 \ meq$.
For $KNO_3$: Molarity $= 0.1 \ M$. Since $1 \ mol$ of $KNO_3$ gives $1 \ mol$ of $NO_3^{-}$,the molarity of $NO_3^{-}$ is $0.1 \ M$. Milliequivalents of $NO_3^{-} = 0.1 \ M \times 200 \ mL = 20 \ meq$.
Total milliequivalents of $NO_3^{-} = 20 + 20 = 40 \ meq$.
Total volume $= 100 \ mL + 200 \ mL = 300 \ mL$.
Normality of $NO_3^{-} = \frac{\text{Total meq}}{\text{Total volume (mL)}} = \frac{40}{300} = 0.133 \ N$.

(C) Mass by mass percentage is the mass of solute dissolved in $100 \ g$ of the solution.
Mass of solution $= 100 \ g$
Mass of glucose $= 10 \ g$
Mass of solvent $= 90 \ g$
Molar mass of glucose $(C_6H_{12}O_6) = 12 \times 6 + 1 \times 12 + 16 \times 6 = 180 \ g \ mol^{-1}$
Molality $= \frac{\text{Mass of solute}}{\text{Molar mass of solute}} \times \frac{1000}{\text{Mass of solvent (in g)}}$
$= \frac{10}{180} \times \frac{1000}{90} = 0.617 \ m$
Volume of solution $= \frac{\text{Mass of solution}}{\text{Density of solution}} = \frac{100 \ g}{1.2 \ g \ mL^{-1}} = 83.33 \ mL$
Molarity $= \frac{\text{Mass of solute}}{\text{Molar mass of solute}} \times \frac{1000}{\text{Volume of solution (in mL)}}$
$= \frac{10}{180} \times \frac{1000}{83.33} = 0.67 \ M$

(A) The number of moles of $X = \frac{W_X}{M_X} = \frac{3.1 \ g}{62 \ g \ mol^{-1}} = 0.05 \ mol$.
The number of moles of $Y = \frac{W_Y}{M_Y} = \frac{19.5 \ g}{78 \ g \ mol^{-1}} = 0.25 \ mol$.
The ratio of mole fractions of $X$ and $Y$ is equal to the ratio of their number of moles:
$\frac{\chi_X}{\chi_Y} = \frac{n_X}{n_Y} = \frac{0.05}{0.25} = \frac{1}{5} = 1:5$.

(D) Molality $(m)$ is defined as the number of moles of solute per kilogram of solvent.
$m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}}$
Since mass is independent of temperature,molality does not change with temperature.
Therefore,the Assertion $(A)$ is incorrect.
The Reason $(R)$ is correct because the expression for molality involves only mass,not volume.

(B) Given:
Density $(d) = 1.12 \ g \ mL^{-1}$
Mass of solute $(w) = 120 \ g$
Molar mass of solute $(M) = 60 \ g \ mol^{-1}$
Mass of solvent $= 1000 \ g$
Total mass of solution $= 1000 \ g + 120 \ g = 1120 \ g$
Using the formula $d = \frac{\text{Mass}}{\text{Volume} (V)}$,we find the volume of the solution:
$V = \frac{1120 \ g}{1.12 \ g \ mL^{-1}} = 1000 \ mL = 1 \ L$
Molarity $(Molarity) = \frac{\text{moles of solute}}{\text{Volume of solution in } L}$
Moles of solute $= \frac{120 \ g}{60 \ g \ mol^{-1}} = 2 \ mol$
Molarity $= \frac{2 \ mol}{1 \ L} = 2.0 \ M$

(D) Let $w_A$ be the mass of water $(H_2O)$ and $w_B$ be the mass of ethanol $(C_2H_5OH)$.
Given: $w_A = x \ g$,$M_A = 18 \ g/mol$,$w_B = 69 \ g$,$M_B = 46 \ g/mol$.
Moles of water $(n_A)$ = $\frac{x}{18}$.
Moles of ethanol $(n_B)$ = $\frac{69}{46} = 1.5 \ mol$.
Mole fraction of ethanol $(X_B)$ = $0.6$.
Since $X_A + X_B = 1$,the mole fraction of water $(X_A)$ = $1 - 0.6 = 0.4$.
Using the formula $X_B = \frac{n_B}{n_A + n_B}$:
$0.6 = \frac{1.5}{\frac{x}{18} + 1.5}$.
$0.6 \times (\frac{x}{18} + 1.5) = 1.5$.
$\frac{0.6x}{18} + 0.9 = 1.5$.
$\frac{x}{30} = 0.6$.
$x = 0.6 \times 30 = 18 \ g$.

(A) Molarity $(M)$ is defined as the number of moles of solute per liter of solution.
Step $1$: Calculate the number of moles of $NaOH$.
Moles of $NaOH = \frac{\text{mass}}{\text{molar mass}} = \frac{0.4 \ g}{40 \ g \ mol^{-1}} = 0.01 \ mol$.
Step $2$: Convert the volume of the solution to liters.
Volume $= 500 \ mL = 0.5 \ L$.
Step $3$: Calculate Molarity.
$M = \frac{\text{moles of solute}}{\text{volume of solution in liters}} = \frac{0.01 \ mol}{0.5 \ L} = 0.02 \ M$.

(C) Initial mass of solution $= 100 \ g$.
Mass of sucrose $= 40 \ g$ and mass of water $= 60 \ g$.
After heating,the mass of sucrose remains constant at $40 \ g$,but the concentration becomes $50 \%$.
Let the new mass of the solution be $m$.
$50 \% \text{ of } m = 40 \ g$.
$m = \frac{40}{0.50} = 80 \ g$.
Mass of water lost $= 100 \ g - 80 \ g = 20 \ g$.

(B) Given: Molality $(m)$ = $1 \ mol \ kg^{-1}$,Density $(d)$ = $1.2 \ g \ mL^{-1}$.
Let the mass of the solvent (water) be $1000 \ g$ $(1 \ kg)$.
The molar mass of glucose $(C_6H_{12}O_6)$ is $180 \ g \ mol^{-1}$.
Since $m = 1 \ mol \ kg^{-1}$,the mass of glucose in $1 \ kg$ of water is $180 \ g$.
Total mass of the solution = Mass of solute + Mass of solvent = $180 \ g + 1000 \ g = 1180 \ g$.
Volume of the solution = $\frac{\text{Mass of solution}}{\text{Density}} = \frac{1180 \ g}{1.2 \ g \ mL^{-1}} = 983.33 \ mL = 0.9833 \ L$.
Molarity $(M)$ = $\frac{\text{Moles of solute}}{\text{Volume of solution in } L} = \frac{1 \ mol}{0.9833 \ L} \approx 1.01 \ M$.

(C) Total mass of the solution $= \text{mass of urea} + \text{mass of glucose} + \text{mass of water} = 6 \ g + 9 \ g + 35 \ g = 50 \ g$.
Mass percent of urea $= \frac{\text{mass of urea}}{\text{total mass of solution}} \times 100 = \frac{6 \ g}{50 \ g} \times 100 = 12 \%$.
Mass percent of glucose $= \frac{\text{mass of glucose}}{\text{total mass of solution}} \times 100 = \frac{9 \ g}{50 \ g} \times 100 = 18 \%$.

(C) Given: Mass of solute (glucose) $m_2 = 18 \ g$,Molar mass of glucose $M_2 = 180 \ g \ mol^{-1}$,Mass of solvent (water) $m_1 = 18 \ g$.
Number of moles of solute $n_2 = \frac{m_2}{M_2} = \frac{18}{180} = 0.1 \ mol$.
Mass of solvent in $kg$ is $m_1 = 18 \ g = 18 \times 10^{-3} \ kg$.
Molality $(m) = \frac{\text{Number of moles of solute}}{\text{Mass of solvent in } kg} = \frac{0.1}{18 \times 10^{-3}} = \frac{0.1}{0.018} = 5.55 \ mol \ kg^{-1}$ or $5.55 \ m$.

(B) The molar mass of $H_2SO_4$ is $2 \times 1.0 + 32.0 + 4 \times 16.0 = 98.0 \ g \ mol^{-1}$.
Number of moles of $H_2SO_4 = \frac{\text{mass}}{\text{molar mass}} = \frac{4.9 \ g}{98.0 \ g \ mol^{-1}} = 0.05 \ mol$.
Volume of solution in liters $= 250 \ mL = 0.250 \ L$.
Molarity $(M) = \frac{\text{moles of solute}}{\text{volume of solution in } L} = \frac{0.05 \ mol}{0.250 \ L} = 0.2 \ mol \ L^{-1}$.

(B) Molarity $(M) = \frac{\text{mass of solute (g)}}{\text{molar mass (g/mol)}} \times \frac{1000}{\text{volume of solution (mL)}}$.
Mass of solution $= \text{mass of solute} + \text{mass of solvent} = 50 \ g + 1000 \ g = 1050 \ g$.
Volume of solution $(V) = \frac{\text{mass of solution}}{\text{density}}$.
At $90^{\circ} C$,$V_1 = \frac{1050}{1.1} \ mL$,so $M_1 = \frac{50}{M_w} \times \frac{1000}{1050/1.1} = \frac{50 \times 1000 \times 1.1}{M_w \times 1050} = \frac{52.38}{M_w}$.
At $10^{\circ} C$,$V_2 = \frac{1050}{1.15} \ mL$,so $M_2 = \frac{50}{M_w} \times \frac{1000}{1050/1.15} = \frac{50 \times 1000 \times 1.15}{M_w \times 1050} = \frac{54.76}{M_w}$.
$\%$ change in molarity $= \frac{M_2 - M_1}{M_1} \times 100 = \frac{54.76/M_w - 52.38/M_w}{52.38/M_w} \times 100 = \frac{2.38}{52.38} \times 100 \approx 4.54 \% \approx 4.5 \%$.

(A) Given: $1 \ M \ Na_2SO_4$ solution means $1 \ mol$ of $Na_2SO_4$ in $1000 \ mL$ of solution.
Density of solution $= 1.2 \ g \ mL^{-1}$.
Mass of $1000 \ mL$ solution $= 1000 \ mL \times 1.2 \ g \ mL^{-1} = 1200 \ g$.
Dissociation reaction: $Na_2SO_4 \rightarrow 2Na^{+} + SO_4^{2-}$.
Since $50 \%$ of $1 \ mol$ of $Na_2SO_4$ dissociates,the amount of $Na_2SO_4$ dissociated $= 0.5 \ mol$.
From the stoichiometry,$1 \ mol$ of $Na_2SO_4$ produces $2 \ mol$ of $Na^{+}$,so $0.5 \ mol$ of $Na_2SO_4$ produces $1 \ mol$ of $Na^{+}$.
Mass of $Na_2SO_4$ remaining $= (1 - 0.5) \times 142 \ g = 71 \ g$.
Mass of solvent $= \text{Mass of solution} - \text{Mass of } Na_2SO_4 = 1200 \ g - 71 \ g = 1129 \ g$.
Molality of $Na^{+}$ ion $= \frac{\text{moles of } Na^{+}}{\text{mass of solvent in } kg} = \frac{1 \ mol}{1.129 \ kg} \approx 0.885 \ m$.
Note: Based on standard calculation,the closest provided option is $0.95 \ m$.

(B) Let the total number of moles in the solution be $1 \ mol$.
Given mole fraction of ethanol $(x_{ethanol})$ $= 0.050$.
Moles of ethanol $(n_{ethanol})$ $= 0.050 \ mol$.
Moles of water $(n_{water})$ $= 1 - 0.050 = 0.950 \ mol$.
Mass of ethanol $= 0.050 \ mol \times 46 \ g/mol = 2.3 \ g$.
Mass of water $= 0.950 \ mol \times 18 \ g/mol = 17.1 \ g$.
Total mass of solution $= 2.3 \ g + 17.1 \ g = 19.4 \ g$.
Assuming the density of the solution is approximately $1 \ g/mL$,the volume of the solution is $19.4 \ mL = 0.0194 \ L$.
Molarity $(M)$ $= \frac{\text{moles of solute}}{\text{volume of solution in } L} = \frac{0.050 \ mol}{0.0194 \ L} \approx 2.58 \ M$.
Rounding to the nearest provided option,the answer is $2.8 \ M$.