Elevation of boiling point of the solvent Questions in English

(A) According to Raoult's law,the addition of a non-volatile solute decreases the vapour pressure of the solvent.
Since the boiling point $(b.p.)$ is the temperature at which the vapour pressure of a liquid becomes equal to the atmospheric pressure,a lower vapour pressure requires a higher temperature to reach atmospheric pressure.
Therefore,the boiling point of the solution increases $(b.p. \text{ elevation})$.
Thus,the correct option is $A$.

(A) According to Raoult's law,when a non-volatile solute is added to a solvent,the vapour pressure of the solution decreases.
Since the boiling point is defined as the temperature at which the vapour pressure of the liquid becomes equal to the atmospheric pressure,a lower vapour pressure requires more heat to reach atmospheric pressure.
Therefore,the boiling point of the solution increases compared to the pure solvent.
Thus,the correct option is $A$.

(A) The ebullioscopic constant $(K_b)$ is given by the formula: $K_b = \frac{R \times T_b^2 \times M}{1000 \times \Delta H_{vap}}$
Where:
$R = 1.987 \, cal \, K^{-1} \, mol^{-1} \approx 2 \, cal \, K^{-1} \, mol^{-1}$
$T_b = 100 + 273 = 373 \, K$
$\Delta H_{vap} = 9700 \, cal/mol$
$M = 18 \, g/mol$ (Molar mass of water)
Substituting the values:
$K_b = \frac{2 \times (373)^2 \times 18}{1000 \times 9700}$
$K_b = \frac{2 \times 139129 \times 18}{9700000}$
$K_b = \frac{5008644}{9700000} \approx 0.516 \, ^oC \, kg/mol$
Thus,the correct option is $A$.

(C) The formula for elevation in boiling point is $\Delta T_b = K_b \times m$.
First,calculate the molality $(m)$ of the solution:
$m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{1.8 \ g / 180 \ g/mol}{100 \ g / 1000 \ g/kg} = \frac{0.01 \ mol}{0.1 \ kg} = 0.1 \ m$.
Now,calculate the molal elevation constant $(K_b)$:
$K_b = \frac{\Delta T_b}{m} = \frac{0.1 \ K}{0.1 \ m} = 1 \ K/m$.

(D) The formula for elevation in boiling point is given by: $\Delta T_b = K_b \times m$,where $m$ is the molality of the solution.
Molality $m = \frac{w \times 1000}{M \times W}$,where $w$ is the mass of solute,$M$ is the molar mass,and $W$ is the mass of solvent in grams.
Substituting the values: $\Delta T_b = \frac{K_b \times w \times 1000}{M \times W}$.
Rearranging for $M$: $M = \frac{K_b \times w \times 1000}{\Delta T_b \times W}$.
Given: $K_b = 2.16 \ ^oC \ kg \ mol^{-1}$,$w = 0.15 \ g$,$W = 15 \ g$,and $\Delta T_b = 0.216 \ ^oC$.
$M = \frac{2.16 \times 0.15 \times 1000}{0.216 \times 15} = \frac{324}{3.24} = 100 \ g \ mol^{-1}$.
Thus,the molecular weight is $100$.

(B) The dissolution of a non-volatile solute in a solvent leads to a decrease in vapor pressure,which results in an increase in the boiling point of the liquid. This phenomenon is known as the elevation of boiling point.

(B) When a non-volatile solute is added to a solvent,the vapour pressure of the solution decreases compared to the pure solvent.
According to the definition of boiling point,it is the temperature at which the vapour pressure of the liquid becomes equal to the atmospheric pressure.
Since the vapour pressure of the solution is lower,it requires more heat to reach the atmospheric pressure.
Therefore,the boiling point of the solution increases. This phenomenon is known as elevation in boiling point.

(B) The formula for elevation in boiling point is $\Delta T_b = \frac{K_b \times w \times 1000}{m \times W}$.
Given: $\Delta T_b = 0.52 \, ^\circ C$,$w = 6 \, g$,$W = 100 \, g$,$K_b = 0.52 \, K \, kg \, mol^{-1}$.
Substituting the values into the formula:
$0.52 = \frac{0.52 \times 6 \times 1000}{m \times 100}$.
$m = \frac{0.52 \times 6 \times 1000}{0.52 \times 100}$.
$m = 60 \, g \, mol^{-1}$.

(B) The elevation in boiling point,denoted as $\Delta T_b$,is defined as the difference between the boiling point of the solution $(T_1)$ and the boiling point of the pure solvent $(T_2)$.
Since the addition of a non-volatile solute increases the boiling point of the solvent,$T_1 > T_2$.
Therefore,the elevation in boiling point is given by $\Delta T_b = T_1 - T_2$.

(B) The elevation in boiling point is given by the formula: $\Delta T_b = K_b \times m$.
Rearranging this equation,we get $K_b = \Delta T_b / m$.
Thus,the molal elevation constant $(K_b)$ is the ratio of the elevation in boiling point $(\Delta T_b)$ to the molality $(m)$ of the solution.

(C) The molality $(m)$ of the solution is calculated as: $m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.1 \, mol}{0.2 \, kg} = 0.5 \, mol \, kg^{-1}$.
Using the formula for elevation in boiling point: $\Delta T_b = K_b \times m$.
$\Delta T_b = 0.513 \, ^oC \, kg \, mol^{-1} \times 0.5 \, mol \, kg^{-1} = 0.2565 \, ^oC$.
The boiling point of the solution $(T_b)$ is: $T_b = T_b^0 + \Delta T_b = 100 \, ^oC + 0.2565 \, ^oC = 100.2565 \, ^oC$.
Rounding to two decimal places,the value is approximately $100.25 \, ^oC$.

(D) The formula for elevation in boiling point is $\Delta T_b = \frac{K_b \times w \times 1000}{M \times W}$.
Here,$\Delta T_b = 1 \ K$,$K_b = 2.53 \ K \ kg \ mol^{-1}$,$w = 10 \ g$,and $W = 100 \ g$.
Rearranging the formula for molar mass $M$: $M = \frac{K_b \times w \times 1000}{\Delta T_b \times W}$.
Substituting the values: $M = \frac{2.53 \times 10 \times 1000}{1 \times 100} = 253 \ g \ mol^{-1}$.

(C) The elevation in boiling point is given by $\Delta T_b = K_b \times m$,where $m$ is the molality of the solution.
For urea: $m_1 = \frac{1 \ g / 60 \ g \cdot mol^{-1}}{V \ L} = \frac{1}{60V} \ mol \cdot L^{-1}$.
For glucose: $m_2 = \frac{3 \ g / 180 \ g \cdot mol^{-1}}{V \ L} = \frac{1}{60V} \ mol \cdot L^{-1}$.
Since the molality $(m)$ is the same for both solutions,the elevation in boiling point $(\Delta T_b)$ will be identical.
Given $\Delta T_b = 100.25 \ ^oC - 100 \ ^oC = 0.25 \ ^oC$ for urea.
Therefore,the boiling point of the glucose solution will be $100 \ ^oC + 0.25 \ ^oC = 100.25 \ ^oC$.

(B) When a non-volatile solute like common salt $(NaCl)$ is dissolved in a solvent like water,the vapor pressure of the solution decreases.
To reach the atmospheric pressure,the solution must be heated to a higher temperature.
Therefore,the boiling point of the solution increases,which is known as the elevation of boiling point.

(B) The elevation in boiling point for a $1 \, \text{molal}$ solution of a solute in a solvent is defined as the molal boiling point elevation constant,also known as the ebullioscopic constant.
It is denoted by $K_b$.
The unit of $K_b$ is $K \, kg \, mol^{-1}$.
For example,for water,$K_b = 0.52 \, K \, kg \, mol^{-1}$.

(D) The elevation in boiling point is given by the formula $\Delta T_b = K_b \times m$,where $K_b$ is the ebullioscopic constant of the solvent and $m$ is the molality of the solution.
Since the molality $(m = 1)$ is constant for all cases,the boiling point elevation depends directly on the value of the ebullioscopic constant $(K_b)$ of the solvent.
The $K_b$ values for the given solvents are approximately:
$1$. Ethyl alcohol: $1.22 \ K \ kg \ mol^{-1}$
$2$. Acetone: $1.71 \ K \ kg \ mol^{-1}$
$3$. Benzene: $2.53 \ K \ kg \ mol^{-1}$
$4$. Chloroform: $3.63 \ K \ kg \ mol^{-1}$
Since Chloroform has the highest $K_b$ value,it will show the maximum boiling point elevation.

(C) The elevation in boiling point is a colligative property,which depends on the van't Hoff factor $(i)$ and the concentration of the solute particles.
The formula is $\Delta T_b = i \times K_b \times m$.
$1$. For $0.1 \ M$ glucose (non-electrolyte),$i = 1$,so effective concentration = $0.1 \times 1 = 0.1 \ M$.
$2$. For $0.2 \ M$ sucrose (non-electrolyte),$i = 1$,so effective concentration = $0.2 \times 1 = 0.2 \ M$.
$3$. For $0.1 \ M$ barium chloride $(BaCl_2 \rightarrow Ba^{2+} + 2Cl^-)$,$i = 3$,so effective concentration = $0.1 \times 3 = 0.3 \ M$.
$4$. For $0.1 \ M$ magnesium sulphate $(MgSO_4 \rightarrow Mg^{2+} + SO_4^{2-})$,$i = 2$,so effective concentration = $0.1 \times 2 = 0.2 \ M$.
Since $BaCl_2$ provides the highest effective concentration of particles $(0.3 \ M)$,it will produce the maximum elevation in boiling point.

(A) The boiling point elevation is a colligative property,which depends on the number of particles in the solution.
The formula for boiling point elevation is $\Delta T_b = i \times K_b \times m$.
Since the mass of the solute is the same $(10 \ g)$,the molality $(m)$ is inversely proportional to the molar mass $(M)$ of the solute.
$NaCl$ $(M \approx 58.5 \ g/mol)$ and $KCl$ $(M \approx 74.5 \ g/mol)$ are electrolytes with a van't Hoff factor $(i)$ of $2$,while sugar ($C_{12}H_{22}O_{11}$,$M \approx 342 \ g/mol$) and glucose ($C_6H_{12}O_6$,$M \approx 180 \ g/mol$) are non-electrolytes $(i = 1)$.
Comparing $NaCl$ and $KCl$,$NaCl$ has a lower molar mass,resulting in a higher molality $(m)$.
Therefore,$NaCl$ produces the highest number of particles per unit mass,leading to the highest boiling point.

(B) The formula for elevation in boiling point is $\Delta T_b = \frac{K_b \times w \times 1000}{M \times W}$.
Given:
$w = 0.1050 \ g$ (mass of solute)
$W = 15.84 \ g$ (mass of solvent)
$\Delta T_b = 100^\circ C$
$K_b = 21.6$ (per $100 \ g$ of solvent,so for $1000 \ g$,$K_b = 216$)
Rearranging for molecular weight $M$:
$M = \frac{K_b \times w \times 1000}{\Delta T_b \times W} = \frac{216 \times 0.1050 \times 1000}{100 \times 15.84} = \frac{22680}{1584} = 143.18 \ g/mol$.

(C) Given: $\Delta T_b = 0.323 \ K$,$w = 0.5143 \ g$ (mass of solute),$W = 35 \ g$ (mass of solvent),${K_b} = 3.9 \ K \ kg \ mol^{-1}$.
Using the formula for elevation in boiling point: $\Delta T_b = \frac{{K_b \times w \times 1000}}{{M_2 \times W}}$.
Rearranging for molecular mass $(M_2)$: $M_2 = \frac{{K_b \times w \times 1000}}{{W \times \Delta T_b}}$.
Substituting the values: $M_2 = \frac{{3.9 \times 0.5143 \times 1000}}{{35 \times 0.323}}$.
$M_2 = \frac{{2005.77}}{{11.305}} = 177.42 \ g/mol$.

(C) Given:
Initial boiling point of water = $100\,^{\circ}C$
Final boiling point of water = $100.52\,^{\circ}C$
Mass of solute $(w)$ = $3\,g$
Mass of solvent $(W)$ = $200\,g$ (assuming density of water is $1\,g/mL$)
${K_b}$ for water = $0.6\,K\,kg\,mol^{-1}$
Elevation in boiling point $(\Delta {T_b})$ = $100.52 - 100 = 0.52\,^{\circ}C$
Using the formula:
$M = \frac{{K_b \times w \times 1000}}{{\Delta {T_b} \times W}}$
$M = \frac{{0.6 \times 3 \times 1000}}{{0.52 \times 200}}$
$M = \frac{{1800}}{{104}} \approx 17.3\,g\,mol^{-1}$

(C) The molality $(m)$ is calculated as: $m = \frac{w_2 \times 1000}{M_2 \times w_1} = \frac{1.8 \times 1000}{180 \times 100} = 0.1 \ m$.
The formula for elevation in boiling point is: $\Delta T_b = i \times K_b \times m$.
For glucose,the van't Hoff factor $(i)$ is $1$.
Given $\Delta T_b = 0.1 \ K$,we have: $0.1 = 1 \times K_b \times 0.1$.
Therefore,$K_b = \frac{0.1}{0.1} = 1 \ \frac{K}{m}$.

(A) Given: Mass of solute $(w)$ = $174.5 \ mg = 0.1745 \ g$,Mass of solvent $(W)$ = $78 \ g$,Molar mass of $S_8$ $(m)$ = $8 \times 32 = 256 \ g \ mol^{-1}$,$K_b = 5.2 \ K \ kg \ mol^{-1}$.
The formula for elevation in boiling point is $\Delta T_b = \frac{1000 \times K_b \times w}{m \times W}$.
Substituting the values: $\Delta T_b = \frac{1000 \times 5.2 \times 0.1745}{256 \times 78}$.
$\Delta T_b = \frac{907.4}{19968} \approx 0.0454 \ K$.
The boiling point of the solution = Boiling point of pure solvent + $\Delta T_b$.
Boiling point = $332.15 + 0.0454 = 332.1954 \ K \approx 332.19 \ K$.

(B) The elevation in boiling point $(\Delta T_b)$ is given by the formula: $\Delta T_b = K_b \times m$,where $K_b$ is the ebullioscopic constant and $m$ is the molality of the solution.
For a $1 \ m$ (one molal) solution,$\Delta T_b = K_b \times 1 = K_b$.
Therefore,the elevation in boiling point for a one molal solution is equal to the ebullioscopic constant $(K_b)$.

(A) The formula for elevation in boiling point is $\Delta T_b = K_b \times m$,where $m$ is the molality.
The molality $m$ is given by $\frac{w \times 1000}{M_2 \times W}$,where $w$ is the mass of solute,$M_2$ is the molar mass of solute,and $W$ is the mass of solvent in grams.
Given: $\Delta T_b = 0.1 \ ^\circ C$,$w = 1.8 \ g$,$M_2 = 180 \ g/mol$ (for glucose),$W = 100 \ g$.
Substituting the values: $0.1 = K_b \times \frac{1.8 \times 1000}{180 \times 100}$.
$0.1 = K_b \times \frac{1800}{18000} = K_b \times 0.1$.
Therefore,$K_b = \frac{0.1}{0.1} = 1 \ K \ kg/mol$.

(C) The boiling point of benzene $(T_b)$ is $80 + 273 = 353\,K$. The latent heat of vaporization $(L_v)$ is $90\,cal/g$.
The elevation in boiling point is $\Delta T_b = 80.175 - 80 = 0.175\,K$.
The molal elevation constant $(K_b)$ is calculated as:
$K_b = \frac{R \times T_b^2}{1000 \times L_v} = \frac{2 \times 353^2}{1000 \times 90} = \frac{2 \times 124609}{90000} \approx 2.769\,K\,kg\,mol^{-1}$.
Using the formula for elevation in boiling point:
$\Delta T_b = \frac{K_b \times w \times 1000}{M \times W}$
where $w = 1\,g$ (solute mass),$W = 83.4\,g$ (solvent mass),and $M$ is the molar mass.
$0.175 = \frac{2.769 \times 1 \times 1000}{M \times 83.4}$
$M = \frac{2769}{0.175 \times 83.4} = \frac{2769}{14.595} \approx 189.72\,g/mol$.
Rounding to the nearest option,the molar mass is $189.79\,g/mol$.

(C) The elevation in boiling point is given by the formula: $\Delta T_b = K_b \times m$
Here,$K_b = 0.513 \ ^oC \ kg \ mol^{-1}$,
Molality $(m)$ = $\frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.1 \ mol}{0.2 \ kg} = 0.5 \ mol \ kg^{-1}$
$\Delta T_b = 0.513 \times 0.5 = 0.2565 \ ^oC$
Since the boiling point of pure water at $1 \ bar$ is $100 \ ^oC$,the boiling point of the solution is $T_b = T_b^0 + \Delta T_b = 100 + 0.2565 = 100.2565 \ ^oC \approx 100.26 \ ^oC$.

(C) The formula for elevation in boiling point is $\Delta T_b = i \times K_b \times m$.
Since glucose is a non-electrolyte,the van't Hoff factor $i = 1$.
The molality $m$ is calculated as: $m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{1.8 / 180}{100 / 1000} = \frac{0.01}{0.1} = 0.1 \ m$.
Substituting the values into the equation: $0.1 = K_b \times 0.1$.
Therefore,$K_b = \frac{0.1}{0.1} = 1 \ \frac{K}{m}$.

(C) Given: Mass of solute $(w)$ = $0.5143 \ g$,Mass of solvent $(W)$ = $35 \ g$,$K_b = 3.9 \ K \ kg \ mol^{-1}$,Elevation in boiling point $(\Delta T_b)$ = $0.323 \ K$.
The formula for elevation in boiling point is: $\Delta T_b = \frac{1000 \times K_b \times w}{M \times W}$.
Rearranging for molar mass $(M)$: $M = \frac{1000 \times K_b \times w}{\Delta T_b \times W}$.
Substituting the values: $M = \frac{1000 \times 3.9 \times 0.5143}{0.323 \times 35}$.
$M = \frac{2005.77}{11.305} \approx 177.42 \ g \ mol^{-1}$.

(B) When a non-volatile solute like salt $(NaCl)$ is added to a solvent like water,the vapor pressure of the solution decreases.
According to the colligative properties,the boiling point of a solution increases (elevation in boiling point) because the vapor pressure of the solution is lower than that of the pure solvent.
Therefore,the boiling point of the solution increases.

(B) The boiling point of a liquid is inversely proportional to its vapor pressure $(T_b \propto 1/P_{vap})$.
When a non-volatile solute is added to a solvent,the vapor pressure of the solution decreases.
To reach the atmospheric pressure,the solution must be heated to a higher temperature.
Therefore,the boiling point of the solution increases,which is known as the elevation of boiling point.

(C) The formula for elevation in boiling point is $\Delta T_b = K_b \times m$,where $m$ is the molality of the solution.
Molality $m = \frac{\text{mass of solute}}{\text{molar mass of solute}} \times \frac{1000}{\text{mass of solvent in } g}$.
Given: mass of solute $= 10 \ g$,molar mass $= 100 \ g/mol$,mass of solvent $= 100 \ g$.
Substituting the values: $m = \frac{10}{100} \times \frac{1000}{100} = 1 \ mol/kg$.
Therefore,$\Delta T_b = K_b \times 1$,which means $\Delta T_b = K_b$.

(B) When a non-volatile solute like common salt $(NaCl)$ is added to a solvent like water,the vapor pressure of the solution decreases.
According to the concept of elevation in boiling point,the boiling point of a solution is higher than that of the pure solvent.
Therefore,the boiling point of the solution increases.

(B) The elevation in boiling point is given by $\Delta T_b = T_b - T_b^\circ = 80.66^\circ \text{C} - 80^\circ \text{C} = 0.66 \ \text{K}$.
The formula for molar mass $(M_2)$ of the solute is:
$M_2 = \frac{K_b \times W_2 \times 1000}{\Delta T_b \times W_1}$
Where:
$K_b = 3.28 \ \text{K kg mol}^{-1}$
$W_2 = 3.3 \ \text{g}$ (mass of solute)
$W_1 = 125 \ \text{g}$ (mass of solvent)
$\Delta T_b = 0.66 \ \text{K}$
Substituting the values:
$M_2 = \frac{3.28 \times 3.3 \times 1000}{0.66 \times 125}$
$M_2 = \frac{10824}{82.5} = 131.2 \ \text{g mol}^{-1}$.

(B) The elevation in boiling point $(\Delta T_b)$ is given by the formula: $\Delta T_b = K_b \times m$,where $K_b$ is the molal elevation constant and $m$ is the molality of the solution.
Therefore,$K_b = \frac{\Delta T_b}{m}$.
Thus,the molal elevation constant is the ratio of the elevation in boiling point to the molality of the solution.

(C) The formula for elevation in boiling point is $\Delta T_b = K_b \times m \times i$.
Since sugar is a non-electrolyte,the van't Hoff factor $i = 1$.
The molality $m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.1 \ mol}{0.2 \ kg} = 0.5 \ mol \ kg^{-1}$.
Therefore,$\Delta T_b = 0.513 \ ^oC \ kg \ mol^{-1} \times 0.5 \ mol \ kg^{-1} = 0.2565 \ ^oC$.
The boiling point of the solution = Boiling point of pure water + $\Delta T_b = 100 \ ^oC + 0.2565 \ ^oC = 100.2565 \ ^oC$.

(B) The formula for molar mass $(M_2)$ using the elevation of boiling point is: $M_2 = \frac{K_b \times w_2 \times 1000}{\Delta T_b \times w_1}$.
Here,$w_2 = 0.1050 \ g$ (mass of solute),$w_1 = 15.84 \ g$ (mass of solvent),$\Delta T_b = 1 \ ^\circ C$ (elevation in boiling point),and $K_b$ per $100 \ g = 21.6$,so $K_b$ per $1000 \ g = 216$.
Substituting the values: $M_2 = \frac{216 \times 0.1050 \times 1000}{1 \times 15.84 \times 10} = \frac{22680}{158.4} \approx 143.18 \ g/mol$.

(B) The elevation in boiling point $(\Delta T_b)$ is defined as the difference between the boiling point of the solution $(T_1)$ and the boiling point of the pure solvent $(T_2)$.
Mathematically,it is expressed as $\Delta T_b = T_1 - T_2$.

(B) The elevation in boiling point is given by the formula: $\Delta T_b = \frac{K_b \times w \times 1000}{M \times W}$
Where:
$K_b = 2.16 \ K \ kg \ mol^{-1}$
$w = 0.11 \ g$ (mass of solute)
$W = 15 \ g$ (mass of solvent)
$\Delta T_b = 0.1 \ ^\circ C$
Rearranging for molecular weight $(M)$:
$M = \frac{K_b \times w \times 1000}{\Delta T_b \times W}$
Substituting the values:
$M = \frac{2.16 \times 0.11 \times 1000}{0.1 \times 15}$
$M = \frac{237.6}{1.5} = 158.4 \ g \ mol^{-1}$
Rounding to the nearest integer,the molecular weight is $158$.

(C) The elevation in boiling point $(\Delta T_b)$ is calculated as:
$\Delta T_b = T_b - T_b^\circ = 354.11 \, K - 353.23 \, K = 0.88 \, K$
The formula for molar mass $(M_2)$ of the solute is:
$M_2 = \frac{K_b \times w_2 \times 1000}{\Delta T_b \times w_1}$
Given:
$K_b = 2.53 \, K \, kg \, mol^{-1}$
$w_2 = 1.80 \, g$
$w_1 = 90 \, g$
$\Delta T_b = 0.88 \, K$
Substituting the values:
$M_2 = \frac{2.53 \times 1.80 \times 1000}{0.88 \times 90} = \frac{4554}{79.2} = 58 \, g \, mol^{-1}$
Thus,the molar mass of the solute is $58 \, g \, mol^{-1}$.

(A) The elevation in boiling point is given by the formula: $\Delta T_b = K_b \times m$.
Here,$\Delta T_b = T_b - T_b^\circ = 100.18 \ ^\circ C - 100 \ ^\circ C = 0.18 \ ^\circ C$ (or $0.18 \ K$).
The molality $(m)$ is $0.1 \ m$.
Substituting the values: $0.18 = K_b \times 0.1$.
Therefore,$K_b = \frac{0.18}{0.1} = 1.8 \ K \ kg \ mol^{-1}$.

(C) The elevation in boiling point is given by the formula $\Delta T_{b} = i K_{b} m$,where $i$ is the van't Hoff factor,$K_{b}$ is the ebullioscopic constant,and $m$ is the molality.
Since the solutions are equimolal ($m$ is constant) and the solvent is the same ($K_{b}$ is constant),the elevation in boiling point depends directly on the van't Hoff factor $i$.
Given that the boiling point of the solution of $X$ is greater than that of $Y$,it implies that $\Delta T_{b}(X) > \Delta T_{b}(Y)$,which means $i_{X} > i_{Y}$.
If $X$ undergoes dissociation,its van't Hoff factor $i$ becomes greater than $1$,leading to a higher boiling point compared to a non-dissociating solute.

(A) The boiling point of water at $1 \ atm$ is $100 \ ^oC$. The elevation in boiling point is $\Delta T_b = 110 \ ^oC - 100 \ ^oC = 10 \ K$.
$CaBr_2$ dissociates as $CaBr_2 \rightarrow Ca^{2+} + 2Br^-$,so the van't Hoff factor $i = 3$.
The initial mass of the solution is $2.8 \ kg = 2800 \ g$. The molality $m = 2 \ mol/kg$. Let $W_s$ be the mass of solute and $W_w$ be the mass of solvent (water).
$m = \frac{n_{solute}}{W_w (in \ kg)}$ $\Rightarrow 2 = \frac{n_{solute}}{W_w}$ $\Rightarrow n_{solute} = 2 W_w$.
Total mass $= W_s + W_w = 2800 \ g$. Molar mass of $CaBr_2 = 40 + 2(80) = 200 \ g/mol$.
$W_s = n_{solute} \times 200 = (2 W_w) \times 200 = 400 W_w$.
$400 W_w + W_w = 2800$ $\Rightarrow 401 W_w = 2800$ $\Rightarrow W_w \approx 6.98 \ kg$ (This implies the initial solution was not $2.8 \ kg$ for $2 \ m$ concentration). Re-evaluating based on the standard formula $\Delta T_b = K_b \times m_{final} \times i$:
$10 = 0.5 \times m_{final} \times 3 \Rightarrow m_{final} = \frac{10}{1.5} = 6.67 \ mol/kg$.
Initial moles of $CaBr_2 = 2 \ mol/kg \times 2 \ kg = 4 \ mol$ (assuming $2 \ kg$ solvent). Mass of $CaBr_2 = 4 \times 200 = 800 \ g$. Initial water $= 2000 \ g$.
$6.67 = \frac{4}{W_{final}} \Rightarrow W_{final} = 0.6 \ kg = 600 \ g$.
Water evaporated $= 2000 \ g - 600 \ g = 1400 \ g$.

(B) The formula for elevation in boiling point is $\Delta T_b = K_b \times m$,where $m$ is the molality of the solution.
Molality $(m)$ is defined as the number of moles of solute per kilogram of solvent.
Moles of solute = $\frac{Y}{M}$.
Mass of solvent in $kg$ = $\frac{250}{1000} = 0.25 \ kg$.
Therefore,$m = \frac{Y / M}{0.25} = \frac{4Y}{M}$.
Substituting this into the elevation formula: $\Delta T_b = K_b \times \frac{4Y}{M} = \frac{4 K_b Y}{M}$.

(C) The ebullioscopic constant $(K_b)$ is related to the thermodynamic properties of the solvent by the formula: $K_b = \frac{R T_b^2 M_1}{1000 \Delta H_{vap}}$.
Comparing this with the given expression $K_b = \frac{m R T_b^2}{1000 X}$,where $m$ is the molar mass of the solvent,we identify $X$ as the latent heat of vaporisation of the solvent $(\Delta H_{vap})$.
Therefore,$X$ represents the latent heat of vaporisation.

(A) The elevation in boiling point is given by the formula: $\Delta T_b = K_b \times m$.
Given: $K_b = 0.52 \ K \ kg \ mol^{-1}$ and molality $m = 0.2 \ m$.
$\Delta T_b = 0.52 \times 0.2 = 0.104 \ ^oC$.
The boiling point of pure water is $100 \ ^oC$.
Therefore,the boiling point of the solution $= 100 + 0.104 = 100.104 \ ^oC$.

(D) The formula for boiling point elevation is $\Delta T_b = K_b \times m$,where $m$ is the molality of the solution.
Molality $m = \frac{w_2 \times 1000}{M_2 \times w_1}$,where $w_2$ is the mass of solute,$M_2$ is the molar mass,and $w_1$ is the mass of solvent in grams.
Substituting the given values: $0.52 = 0.52 \times \frac{12.5 \times 1000}{M_2 \times 250}$.
Simplifying the equation: $1 = \frac{12500}{M_2 \times 250}$.
$M_2 = \frac{12500}{250} = 50 \ g \ mol^{-1}$.

(C) The elevation in boiling point $(\Delta T_b)$ is a colligative property given by the formula $\Delta T_b = i \times K_b \times m$,where $i$ is the van't Hoff factor,$K_b$ is the ebullioscopic constant,and $m$ is the molality of the solution.
For all three solutions,the solvent is $1 \ kg$ of $H_2O$ and the amount of solute is $1 \ mole$.
Thus,the molality $(m)$ is the same for all three solutions $(m = 1 \ mol / 1 \ kg = 1 \ m)$.
Glucose,urea,and the non-electrolyte non-volatile unknown solute $A$ are all non-electrolytes,so their van't Hoff factor $(i)$ is $1$ for all of them.
Since $i$,$K_b$,and $m$ are identical for all three solutions,the elevation in boiling point $(\Delta T_b)$ will be the same.
Therefore,the boiling point of the solutions will be equal: $I = II = III$.

(A) The molar mass of $CaCl_2$ is $40 + 2 \times 35.5 = 111 \ g \ mol^{-1}$.
The number of moles of $CaCl_2$ is $n = \frac{11.1 \ g}{111 \ g \ mol^{-1}} = 0.1 \ mol$.
Since $CaCl_2$ dissociates as $CaCl_2 \rightarrow Ca^{2+} + 2Cl^-$,the van't Hoff factor $i = 3$.
The elevation in boiling point is given by $\Delta T_b = i \times K_b \times m$,where $m$ is the molality.
$m = \frac{0.1 \ mol}{1 \ kg} = 0.1 \ mol \ kg^{-1}$.
$\Delta T_b = 3 \times 0.5 \ K \ kg \ mol^{-1} \times 0.1 \ mol \ kg^{-1} = 0.15 \ K$.

(A) The ebullioscopic constant $(K_b)$ is a characteristic property of the solvent.
It depends only on the nature of the solvent and is independent of the solute dissolved in it.
Since the solvent in both cases is water,the value of $K_b$ will remain the same,i.e.,$0.51 \ K \ kg \ mol^{-1}$.