Elevation of boiling point of the solvent Questions in English

(C) The elevation in boiling point 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.
Since the mass of the solute $(1 \ g)$ and the mass of the solvent $(100 \ g)$ are the same for both solvents $A$ and $B$,the molality $m$ is the same for both solutions $(m_A = m_B)$.
Therefore,the ratio of the elevation in boiling points is directly proportional to the ratio of their ebullioscopic constants:
$\frac{\Delta T_b (A)}{\Delta T_b (B)} = \frac{K_{b(A)}}{K_{b(B)}}$.
Given that the ratio of the ebullioscopic constants is $\frac{K_{b(A)}}{K_{b(B)}} = \frac{1}{5}$,the ratio of the elevation in boiling points is $1:5$.

(A) Step $1$: Calculate the number of moles of $CuCl_2$: $n = \frac{13.44 \ g}{134.4 \ g \ mol^{-1}} = 0.1 \ mol$.
Step $2$: Calculate molality $(m)$: $m = \frac{0.1 \ mol}{1 \ kg} = 0.1 \ mol \ kg^{-1}$.
Step $3$: Determine the van't Hoff factor $(i)$: $CuCl_2$ dissociates as $CuCl_2 \rightarrow Cu^{2+} + 2Cl^-$. Assuming $100 \%$ ionization,$i = 3$.
Step $4$: Calculate elevation in boiling point: $\Delta T_b = i \times K_b \times m = 3 \times 0.52 \times 0.1 = 0.156 \ K \approx 0.16 \ K$.

(A) The elevation in boiling point is given by the formula $\Delta T_b = i \times K_b \times m$,where $i$ is the van't Hoff factor and $m$ is the molality.
Since $K_b$ is constant for a given solvent,the boiling point depends on the product $i \times m$.
For $0.01 \ M \ Na_2SO_4$,$i = 3$ $(Na_2SO_4 \rightarrow 2Na^+ + SO_4^{2-})$,so $i \times m = 3 \times 0.01 = 0.03$.
For $0.01 \ M \ KNO_3$,$i = 2$ $(KNO_3 \rightarrow K^+ + NO_3^-)$,so $i \times m = 2 \times 0.01 = 0.02$.
For $0.015 \ M \ \text{urea}$,$i = 1$ (non-electrolyte),so $i \times m = 1 \times 0.015 = 0.015$.
For $0.015 \ M \ \text{glucose}$,$i = 1$ (non-electrolyte),so $i \times m = 1 \times 0.015 = 0.015$.
Since $0.01 \ M \ Na_2SO_4$ has the highest value of $i \times m$,it will exhibit the highest boiling point.

(A) The elevation in boiling point is given by the formula $\Delta T_b = i \times K_b \times m$.
Since $K_b$ is constant for the solvent,$\Delta T_b$ depends on the product of the van't Hoff factor $(i)$ and molality $(m)$.
For $0.01 \ M \ Na_2SO_4$,$i = 3$ (as it dissociates into $2Na^+ + SO_4^{2-}$),so $i \times m = 3 \times 0.01 = 0.03$.
For $0.01 \ M \ KNO_3$,$i = 2$ (as it dissociates into $K^+ + NO_3^-$),so $i \times m = 2 \times 0.01 = 0.02$.
For $0.015 \ M$ urea,$i = 1$ (non-electrolyte),so $i \times m = 1 \times 0.015 = 0.015$.
For $0.015 \ M$ glucose,$i = 1$ (non-electrolyte),so $i \times m = 1 \times 0.015 = 0.015$.
Since $0.01 \ M \ Na_2SO_4$ has the highest value of $i \times m$,it will exhibit the highest boiling point.

(C) Sucrose is a non-electrolyte,so the van't Hoff factor $i = 1$.
The boiling point elevation is given by the formula: $\Delta T_b = i K_b m$.
For the first solution $(m_1 = 1 \ m)$: $\Delta T_{b_1} = T_1 - T_0 = 1 \times 0.52 \times 1 = 0.52 \ K$.
For the second solution $(m_2 = 2 \ m)$: $\Delta T_{b_2} = T_2 - T_0 = 1 \times 0.52 \times 2 = 1.04 \ K$.
Subtracting the two equations: $(T_2 - T_0) - (T_1 - T_0) = 1.04 - 0.52$.
Therefore,$T_2 - T_1 = 0.52 \ K$.

(C) The formula for the molar mass $(M_2)$ of a solute is given by:
$M_2 = \frac{1000 \times K_b \times w_2}{\Delta T_b \times w_1}$
Where:
$w_2 = 2.5 \ g$ (mass of solute)
$w_1 = 34 \ g$ (mass of solvent)
$\Delta T_b = 1.38 \ ^\circ C$ (elevation in boiling point)
$K_b = 2.53 \ ^\circ C \ m^{-1}$ (ebullioscopic constant)
Substituting the values into the formula:
$M_2 = \frac{1000 \times 2.53 \times 2.5}{1.38 \times 34}$
This can be written as:
$2.5 \times 10^3 \times \frac{2.53}{34} \times \frac{1}{1.38}$

(D) The elevation in boiling point is given by the formula $\Delta T_b = K_b \times m$,where $m$ is the molality of the solution.
For the first solution: $0.51 = K_b \times 0.1$,so $K_b = 5.1 \ K \cdot kg \cdot mol^{-1}$.
When we mix two solutions of the same molality $(0.1 \ m)$,the resulting solution will also have a molality of $0.1 \ m$ because molality is an intensive property.
Since the molality $(m)$ of the resulting solution remains $0.1 \ m$,the elevation in boiling point $\Delta T_b$ will remain $K_b \times 0.1 = 5.1 \times 0.1 = 0.51 \ K$.

(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 $m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}}$.
Moles of solute = $\frac{\text{mass of solute}}{\text{molar mass of solute}} = \frac{5}{M_A}$.
Mass of solvent = $200 \ g = 0.2 \ kg$.
Therefore,$m = \frac{5 / M_A}{0.2} = \frac{5}{0.2 \times M_A} = \frac{5}{M_A / 5} = \frac{25}{M_A}$.
Substituting this into the elevation formula: $\Delta T_b = K_b \times \frac{25}{M_A} = \frac{25 K_b}{M_A}$.

(C) The elevation of 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,the molal elevation constant $K_b$ is the ratio of the elevation of boiling point $(\Delta T_b)$ to the molality $(m)$: $K_b = \frac{\Delta T_b}{m}$.

(A) The molal elevation constant $(K_b)$ is related to the enthalpy of vaporization $(\Delta_{vap}H)$ by the formula: $K_b = \frac{R \times T_b^2 \times M}{1000 \times \Delta_{vap}H}$.
Here,$T_b = 110.7 273.15 = 383.85\, K$.
$R = 8.314\, J\, K^{-1}\, mol^{-1}$.
$M$ (molar mass of toluene,$C_7H_8$) = $92\, g\, mol^{-1}$.
Rearranging for $\Delta_{vap}H$: $\Delta_{vap}H = \frac{R \times T_b^2 \times M}{1000 \times K_b}$.
$\Delta_{vap}H = \frac{8.314 \times (383.85)^2 \times 92}{1000 \times 3.32}$.
$\Delta_{vap}H = \frac{8.314 \times 147340.82 \times 92}{3320} \approx 33945\, J\, mol^{-1} \approx 34\, kJ\, mol^{-1}$.

(B) The formula for the elevation of boiling point 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 of solute,and $w_1$ is the mass of solvent in grams.
Given: $\Delta T_b = 0.3 \ K$,$w_2 = 0.5 \ g$,$w_1 = 35 \ g$,$K_b = 3.9 \ K \ kg \ mol^{-1}$.
Substituting the values: $0.3 = \frac{3.9 \times 0.5 \times 1000}{M_2 \times 35}$.
$M_2 = \frac{3.9 \times 0.5 \times 1000}{0.3 \times 35} = \frac{1950}{10.5} \approx 185.7 \ g \ mol^{-1}$.

(B) The formula for elevation in boiling point is $\Delta T_b = K_b \times m$,where $m$ is the molality.
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 of solute,and $w_1$ is the mass of solvent in grams.
Given: $\Delta T_b = 0.215 \ K$,$w_2 = 0.15 \ g$,$w_1 = 15 \ g$,$K_b = 2.15 \ K \ kg \ mol^{-1}$.
Substituting the values: $0.215 = \frac{2.15 \times 0.15 \times 1000}{M_2 \times 15}$.
$0.215 = \frac{2.15 \times 150}{M_2 \times 15}$.
$0.215 = \frac{2.15 \times 10}{M_2}$.
$M_2 = \frac{21.5}{0.215} = 100 \ g \ mol^{-1}$.

(A) The ebullioscopic constant $(K_b)$ is given by the formula: $K_b = \frac{R \cdot M_1 \cdot T_b^2}{1000 \cdot \Delta H_{vap}}$.
Here,$R$ is the gas constant,$M_1$ is the molar mass of the solvent,$T_b$ is the boiling point of the solvent,and $\Delta H_{vap}$ is the enthalpy of vaporization.
Given that $\Delta H_{vap}$ is approximately the same for all,$K_b$ is directly proportional to $M_1 \cdot T_b^2$.
Comparing the molar masses $(M_1)$ and boiling points $(T_b)$:
$1$. Naphthalene $(C_{10}H_8)$: $M_1 = 128 \ g/mol$,$T_b = 491 \ K$
$2$. Toluene $(C_7H_8)$: $M_1 = 92 \ g/mol$,$T_b = 384 \ K$
$3$. Benzene $(C_6H_6)$: $M_1 = 78 \ g/mol$,$T_b = 353 \ K$
$4$. Water $(H_2O)$: $M_1 = 18 \ g/mol$,$T_b = 373 \ K$
Calculating the product $M_1 \cdot T_b^2$ shows that Naphthalene has the highest value due to its high molar mass and high boiling point.

(D) Given: Mass of solute $(w_2)$ = $3 \ g$, Volume of water = $200 \ mL$, so Mass of solvent $(w_1)$ = $200 \ g = 0.2 \ kg$.
Boiling point of solution $(T_b)$ = $100.52 \ ^oC$, Boiling point of pure water $(T_b^o)$ = $100 \ ^oC$.
Elevation in boiling point $(\Delta T_b)$ = $T_b - T_b^o = 100.52 - 100 = 0.52 \ K$.
Formula: $\Delta T_b = K_b \times m = K_b \times \frac{w_2 \times 1000}{M_2 \times w_1}$.
Substituting values: $0.52 = 0.6 \times \frac{3 \times 1000}{M_2 \times 200}$.
$0.52 = 0.6 \times \frac{15}{M_2}$.
$M_2 = \frac{0.6 \times 15}{0.52} = \frac{9}{0.52} \approx 17.3 \ g \ mol^{-1}$.

(B) Weight of solute $(w) = 1 \ g$
Weight of solvent $(W) = 75 \ g$
Boiling point of solution $= 100.114 \ ^oC$
Boiling point of solvent $= 100 \ ^oC$
$\Delta T = 100.114 - 100 = 0.114 \ ^oC$
Molecular weight of solute $(m) = 60.1$
Boiling point elevation constant $(K_b) = ?$
The formula for boiling point elevation is $\Delta T = K_b \times m_{olality} = K_b \times \frac{w \times 1000}{m \times W}$
Rearranging for $K_b$: $K_b = \frac{\Delta T \times m \times W}{w \times 1000}$
$K_b = \frac{0.114 \times 60.1 \times 75}{1 \times 1000}$
$K_b = \frac{513.855}{1000} \approx 0.513 \ ^oC \ kg \ mol^{-1}$

(A) $1$. Calculate the moles of glucose: $\text{Moles} = \frac{18 \ g}{180 \ g \ mol^{-1}} = 0.1 \ mol$.
$2$. Determine the molality $(m)$ of the solution: $m = \frac{0.1 \ mol}{1 \ kg} = 0.1 \ mol \ kg^{-1}$.
$3$. Calculate the elevation in boiling point $(\Delta T_b)$: $\Delta T_b = K_b \times m = 0.52 \ K \ kg \ mol^{-1} \times 0.1 \ mol \ kg^{-1} = 0.052 \ K$.
$4$. Determine the boiling point of the solution: $T_b = T_b^0 + \Delta T_b = 373.15 \ K + 0.052 \ K = 373.202 \ K$.

(N/A) The elevation in boiling point $(\Delta T_{b})$ is given by: $\Delta T_{b} = T_{b} - T_{b}^{\circ} = 354.11 \, K - 353.23 \, K = 0.88 \, K$.
The formula for molar mass $(M_{2})$ of the solute is: $M_{2} = \frac{K_{b} \times w_{2} \times 1000}{\Delta T_{b} \times w_{1}}$.
Substituting the given values: $M_{2} = \frac{2.53 \, K \, kg \, mol^{-1} \times 1.80 \, g \times 1000 \, g \, kg^{-1}}{0.88 \, K \times 90 \, g} = 58 \, g \, mol^{-1}$.
Thus,the molar mass of the solute is $58 \, g \, mol^{-1}$.

(D) The elevation in boiling point is $\Delta T_{b} = 100^{\circ} \, C - 99.63^{\circ} \, C = 0.37 \, K$.
Mass of solvent $(w_{1}) = 500 \, g$.
Molar mass of sucrose $(C_{12}H_{22}O_{11}) (M_{2}) = 342 \, g \, mol^{-1}$.
Molal elevation constant $(K_{b})$ for water is $0.52 \, K \, kg \, mol^{-1}$.
Using the formula $\Delta T_{b} = \frac{K_{b} \times 1000 \times w_{2}}{M_{2} \times w_{1}}$,we get:
$w_{2} = \frac{\Delta T_{b} \times M_{2} \times w_{1}}{K_{b} \times 1000} = \frac{0.37 \times 342 \times 500}{0.52 \times 1000} \approx 121.67 \, g$.
Thus,$121.67 \, g$ of sucrose must be added.

(N/A) The boiling point of a liquid is the temperature at which its vapour pressure becomes equal to the atmospheric pressure $(1.013 \ bar)$.
When a non-volatile solute is added to a pure solvent,the vapour pressure of the resulting solution decreases compared to the pure solvent at the same temperature.
Because the vapour pressure of the solution is lower,it must be heated to a higher temperature to reach the atmospheric pressure $(1.013 \ bar)$.
This increase in the boiling point of the solution compared to the pure solvent is known as the elevation in boiling point,denoted by $\Delta T_b$.

(N/A) The elevation in boiling point is defined as the increase in the boiling point of a solution when $1 \ mol$ of a non-volatile solute is dissolved in $1 \ kg$ of solvent.
The boiling point of a solution is always higher than that of the pure solvent. This elevation depends on the number of solute particles rather than their nature.
Let $T_{b}^{0}$ be the boiling point of the pure solvent and $T_{b}$ be the boiling point of the solution. The increase in boiling point is given by $\Delta T_{b} = T_{b} - T_{b}^{0}$.
For dilute solutions,the elevation of boiling point $(\Delta T_{b})$ is directly proportional to the molal concentration $(m)$ of the solute:
$\Delta T_{b} \propto m$
$\Delta T_{b} = K_{b} \times m$
Where $K_{b}$ is the Boiling Point Elevation Constant or Ebullioscopic Constant. Its unit is $K \ kg \ mol^{-1}$.
If $w_{2}$ grams of solute with molar mass $M_{2}$ are dissolved in $w_{1}$ grams of solvent,the molality $(m)$ is:
$m = \frac{w_{2} \times 1000}{M_{2} \times w_{1}}$
Substituting $m$ into the equation for $\Delta T_{b}$:
$\Delta T_{b} = \frac{K_{b} \times 1000 \times w_{2}}{M_{2} \times w_{1}}$
Rearranging to find the molar mass of the solute:
$M_{2} = \frac{1000 \times w_{2} \times K_{b}}{\Delta T_{b} \times w_{1}}$

(A) The molal elevation constant $(K_b)$ is defined by the equation: $\Delta T_b = K_b \times m$,where $\Delta T_b$ is the elevation in boiling point $(K)$ and $m$ is the molality $(mol \ kg^{-1})$.
Rearranging for $K_b$: $K_b = \frac{\Delta T_b}{m} = \frac{K}{mol \ kg^{-1}} = K \ kg \ mol^{-1}$.
Thus,the unit is $K \ kg \ mol^{-1}$.

(A) The boiling point elevation constant $(K_b)$,also known as the ebullioscopic constant,is defined by the formula:
$K_b = \frac{R \times M_1 \times T_b^2}{1000 \times \Delta_{vap}H}$
Where:
$R$ is the universal gas constant,
$M_1$ is the molar mass of the solvent,
$T_b$ is the boiling point of the pure solvent in Kelvin,
$\Delta_{vap}H$ is the enthalpy of vaporization of the solvent.

(N/A) The boiling point of a liquid depends on its vapor pressure relative to the atmospheric pressure. At a fixed atmospheric pressure,a lower vapor pressure leads to a higher boiling point,and vice-versa.
$NaCl$ is a non-volatile solute. When added to water,it occupies the surface area and reduces the escaping tendency of water molecules,thereby lowering the vapor pressure of the solution. Consequently,the boiling point of water increases.
Conversely,methyl alcohol $(CH_3OH)$ is more volatile than water. Its addition increases the total vapor pressure of the solution,which leads to a decrease in the boiling point of the resulting mixture.

(D) The elevation in boiling point is given by the formula: $\Delta T_{b} = i \times K_{b} \times m$.
For $NaCl$,the van't Hoff factor $i = 2$.
The molality $m$ is defined as: $m = \frac{W_{NaCl} \times 1000}{M_{NaCl} \times W_{H_2O(g)}}$.
Given $\Delta T_{b} = 1 \,K$,$K_{b} = 0.52 \,K \cdot kg/mol$,$M_{NaCl} = 58.5 \,g/mol$,and $W_{H_2O} = 500 \,g$.
Substituting the values: $1 = 2 \times 0.52 \times \frac{W_{NaCl} \times 1000}{58.5 \times 500}$.
Solving for $W_{NaCl}$: $W_{NaCl} = \frac{1 \times 58.5 \times 500}{2 \times 0.52 \times 1000} = 28.125 \,g$.

(A) The formula for elevation in boiling point is $\Delta T_b = K_b \times m$,where $m$ is the molality of the solution.
Molality $m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{3.00 / M}{100 / 1000} = \frac{30}{M} \, mol \, kg^{-1}$.
Given $\Delta T_b = 0.60 \, K$ and $K_b = 5.0 \, K \, kg \, mol^{-1}$.
Substituting the values: $0.60 = 5.0 \times \left( \frac{30}{M} \right)$.
$0.60 = \frac{150}{M}$.
$M = \frac{150}{0.60} = 250 \, g \, mol^{-1}$.

(D) Given:
Mass of solute $(w_2)$ $= 2.5 \times 10^{-3} \ kg = 2.5 \ g$
Mass of solvent $(w_1)$ $= 75 \times 10^{-3} \ kg = 75 \ g$
$K_b = 0.52 \ K \ kg \ mol^{-1}$
Boiling point of solution $(T_b)$ $= 373.535 \ K$
Boiling point of pure water $(T_b^o)$ $= 373.15 \ K$
Elevation in boiling point $(\Delta T_b)$ $= T_b - T_b^o = 373.535 - 373.15 = 0.385 \ K$
Formula: $\Delta T_b = K_b \times \frac{w_2 \times 1000}{M_2 \times w_1}$
$0.385 = 0.52 \times \frac{2.5 \times 1000}{M_2 \times 75}$
$M_2 = \frac{0.52 \times 2500}{0.385 \times 75} = \frac{1300}{28.875} \approx 45.02 \ g \ mol^{-1}$
Rounding to the nearest integer,the molar mass is $45 \ g \ mol^{-1}$.

(A) The elevation in boiling point is given by the formula $\Delta T_b = K_b \cdot m$,where $K_b$ is the ebullioscopic constant and $m$ is the molality of the solution.
Since the mass of the solute and the mass of the solvent are the same for both solutions,the molality $m$ is identical for both solvents $A$ and $B$.
Given the ratio of ebullioscopic constants $\frac{(K_b)_A}{(K_b)_B} = \frac{1}{8}$.
Therefore,the ratio of the elevation in boiling points is $\frac{(\Delta T_b)_A}{(\Delta T_b)_B} = \frac{(K_b)_A \cdot m}{(K_b)_B \cdot m} = \frac{(K_b)_A}{(K_b)_B} = \frac{1}{8}$.
Comparing this to $\frac{x}{y}$,we get $x = 1$ and $y = 8$.
The value of $y$ is $8$.

(B) For a dilute aqueous solution,the elevation in boiling point is given by $\Delta T_B = K_B \times m$,where $m$ is the molality.
Since the boiling points are equal,the elevation in boiling point is the same,so the molalities must be equal: $m_A = m_B$.
For $A$: $2 \ g$ of solute in $98 \ g$ of solvent. Molality $m_A = \frac{2 / M_A}{98 / 1000} = \frac{2000}{98 M_A}$.
For $B$: $8 \ g$ of solute in $92 \ g$ of solvent. Molality $m_B = \frac{8 / M_B}{92 / 1000} = \frac{8000}{92 M_B}$.
Equating $m_A = m_B$: $\frac{2000}{98 M_A} = \frac{8000}{92 M_B}$.
$\frac{1}{98 M_A} = \frac{4}{92 M_B}$.
$M_B = \frac{4 \times 98}{92} M_A = \frac{392}{92} M_A \approx 4.26 M_A$.
Given the standard approximation often used in such problems where the mass of solvent is taken as $100 \ g$ (assuming dilute solution),the ratio becomes $\frac{2/M_A}{100} = \frac{8/M_B}{100}$,which leads to $M_B = 4M_A$.

(C) Given,mass of glucose $(C_{6}H_{12}O_{6})$,$w_{2} = 18 \, g$.
Mass of solvent (water),$w_{1} = 1 \, kg$.
$K_{b}$ of water $= 0.52 \, K \, kg \, mol^{-1}$.
Molar mass of glucose $(M_{2}) = 6 \times 12 + 12 \times 1 + 6 \times 16 = 180 \, g \, mol^{-1}$.
Number of moles of glucose $(n_{2}) = \frac{18 \, g}{180 \, g \, mol^{-1}} = 0.1 \, mol$.
Molality $(m) = \frac{n_{2}}{w_{1} \text{ (in kg)}} = \frac{0.1 \, mol}{1 \, kg} = 0.1 \, m$.
Elevation in boiling point,$\Delta T_{b} = K_{b} \times m = 0.52 \times 0.1 = 0.052 \, K$.
Boiling point of solution,$T_{b} = T_{b}^{\circ} + \Delta T_{b} = 373.15 \, K + 0.052 \, K = 373.202 \, K$.
Thus,the boiling point is closest to $373.20 \, K$.

(B) The elevation in boiling point is given by $\Delta T_b = T_b - T_b^\circ = 373.52 \ K - 373 \ K = 0.52 \ K$.
Using the formula $\Delta T_b = K_b \times m$,where $m$ is the molality of the solution.
Molality $m = \frac{\text{mass of solute}}{\text{molar mass of solute}} \times \frac{1000}{\text{mass of solvent in } g}$.
Substituting the values: $0.52 = 0.52 \times \frac{2}{M} \times \frac{1000}{20}$.
$1 = \frac{2}{M} \times 50$.
$M = 100 \ g \ mol^{-1}$.

(B) The boiling point of a liquid is the temperature at which its vapour pressure becomes equal to the external atmospheric pressure $(1 \ atm)$.
Looking at the provided graph,the curve for the solvent intersects the $1 \ atm$ line at a temperature of $82^{\circ} C$.
Therefore,the boiling point of the solvent is $82^{\circ} C$.

(A) The boiling point elevation constant $K_B$ is given by the formula: $K_B = \frac{R \cdot T_B^2 \cdot M}{1000 \cdot \Delta H_{vap}}$.
Since the molecular weights $M$ are the same and $R$ is a constant,we have $K_B \propto \frac{T_B^2}{\Delta H_{vap}}$.
Given $\frac{(T_B)_X}{(T_B)_Y} = \frac{2}{1}$ and $\frac{(\Delta H)_X}{(\Delta H)_Y} = \frac{1}{2}$.
Therefore,$\frac{(K_B)_X}{(K_B)_Y} = \left( \frac{(T_B)_X}{(T_B)_Y} \right)^2 \times \frac{(\Delta H)_Y}{(\Delta H)_X}$.
Substituting the values: $\frac{(K_B)_X}{(K_B)_Y} = (2)^2 \times \left( \frac{2}{1} \right) = 4 \times 2 = 8$.
Thus,$m = 8$.

(D) Mass of solution = $100 \, g$.
Mass of $NaCl = 29.25 \, g$,Mass of $MgCl_2 = 19 \, g$.
Mass of solvent $(H_2O)$ = $100 - (29.25 + 19) = 51.75 \, g = 0.05175 \, kg$.
For $NaCl$ $(i=2)$: Moles = $29.25 / 58.5 = 0.5 \, mol$.
For $MgCl_2$ $(i=3)$: Moles = $19 / 95 = 0.2 \, mol$.
Total moles of ions = $(i_{NaCl} \times n_{NaCl}) + (i_{MgCl_2} \times n_{MgCl_2}) = (2 \times 0.5) + (3 \times 0.2) = 1.0 + 0.6 = 1.6 \, mol$.
$\Delta T_b = i \times K_b \times m = K_b \times (n_{total_ions} / \text{mass of solvent in kg}) = 0.52 \times (1.6 / 0.05175) \approx 16.07 \, ^{\circ}C$.
Boiling point of solution = $100 + 16.07 = 116.07 \, ^{\circ}C$.
Nearest integer is $116$.

(B) From the graph,the boiling points of pure solvents $X$ and $Y$ are $360 \ K$ and $367 \ K$ respectively,and for their $NaCl$ solutions are $362 \ K$ and $368 \ K$ respectively.
For $NaCl$ (completely dissociated,$i = 2$):
$(\Delta T_b)_X = i (K_b)_X m = 362 - 360 = 2 \ K$
$(\Delta T_b)_Y = i (K_b)_Y m = 368 - 367 = 1 \ K$
Dividing the two equations: $\frac{(K_b)_X}{(K_b)_Y} = \frac{2}{1} = 2$.
For solute $S$ undergoing dimerization,the Van't Hoff factor is $i = 1 - \frac{\alpha}{2}$.
Given $\alpha_Y = 0.7$,so $i_Y = 1 - \frac{0.7}{2} = 0.65$.
Given $(\Delta T_b)_X = 3 (\Delta T_b)_Y$ for solute $S$:
$i_X (K_b)_X m = 3 \cdot i_Y (K_b)_Y m$
$(1 - \frac{\alpha_X}{2}) \cdot (K_b)_X = 3 \cdot (0.65) \cdot (K_b)_Y$
$(1 - \frac{\alpha_X}{2}) \cdot 2 = 1.95$
$1 - \frac{\alpha_X}{2} = 0.975$
$\frac{\alpha_X}{2} = 0.025$
$\alpha_X = 0.05$.

(A) The elevation in boiling point is given by $\Delta T_b = i \times K_b \times m$,where $m$ is the molality.
For Vessel-$1$: $(\Delta T_b)_1 = i_1 \times K_b \times \frac{w_2 / GMM_X}{w_1 / 1000}$.
For Vessel-$2$: $(\Delta T_b)_2 = i_2 \times K_b \times \frac{w_2 / GMM_Y}{w_1 / 1000}$.
Taking the ratio: $\frac{(\Delta T_b)_1}{(\Delta T_b)_2} = \frac{i_1}{i_2} \times \frac{GMM_Y}{GMM_X}$.
Given $GMM_X = 0.8 \times GMM_Y$ and $i_1 = 1.2 \times i_2$.
Substituting these values: $\frac{(\Delta T_b)_1}{(\Delta T_b)_2} = \frac{1.2 \times i_2}{i_2} \times \frac{GMM_Y}{0.8 \times GMM_Y} = \frac{1.2}{0.8} = 1.5$.
Therefore,the percentage is $1.5 \times 100 = 150 \%$.

(A) The elevation in boiling point is given by $\Delta T_{b} = i \cdot K_{b} \cdot m$,where $i$ is the van't Hoff factor and $m$ is the molality (proportional to concentration $C$).
Since $K_{b}$ is constant for the solvent,$\Delta T_{b} \propto i \cdot C$.

Solution	$i \cdot C$
$(i) \ 10^{-4} \ M \ NaCl$	$2 \times 10^{-4}$
$(ii) \ 10^{-4} \ M \ \text{Urea}$	$1 \times 10^{-4}$
$(iii) \ 10^{-3} \ M \ NaCl$	$2 \times 10^{-3}$
$(iv) \ 10^{-2} \ M \ NaCl$	$2 \times 10^{-2}$

Comparing the values of $i \cdot C$: $1 \times 10^{-4} < 2 \times 10^{-4} < 2 \times 10^{-3} < 2 \times 10^{-2}$.
Therefore,the order of increasing boiling points is $(ii) < (i) < (iii) < (iv)$.

(A) The molality $(m)$ of the solution is calculated as: $m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}}$.
Given mass of glucose $= 36 \ g$,molar mass $= 180 \ g/mol$.
Moles of glucose $= \frac{36}{180} = 0.2 \ mol$.
Assuming the density of the solution is approximately $1 \ g/mL$,$1 \ dm^3$ (or $1 \ L$) of solution contains $1000 \ g$ of water,which is $1 \ kg$.
Thus,$m = 0.2 \ mol / 1 \ kg = 0.2 \ m$.
The elevation in boiling point is given by $\Delta T_{b} = K_{b} \times m = K_{b} \times 0.2 = 0.2 K_{b}$.
The boiling point of the solution $= 100 + \Delta T_{b} = 100 + 0.2 K_{b}$.

(D) Since both solutions boil at the same temperature,their elevation in boiling point $(\Delta T_b)$ must be the same.
Since $\Delta T_b = K_b \times m$,and $K_b$ is the same for the same solvent (water),the molality $(m)$ of both solutions must be equal.
Given concentration in $g \ dm^{-3}$ is $C = \frac{W}{V}$. For a fixed volume of solvent,the molarity and molality are proportional to the concentration divided by the molar mass.
$\frac{C_{\text{glucose}}}{M_{\text{glucose}}} = \frac{C_A}{M_A}$
Substituting the given values:
$\frac{18}{180} = \frac{6}{M_A}$
$0.1 = \frac{6}{M_A}$
$M_A = \frac{6}{0.1} = 60 \ g \ mol^{-1}$.

(C) The elevation in boiling point is given by the formula: $\Delta T_b = T_b - T_b^\circ$
Given,$T_b = 319.8 \ K$ and $T_b^\circ = 319.5 \ K$.
So,$\Delta T_b = 319.8 \ K - 319.5 \ K = 0.3 \ K$.
The formula for elevation in boiling point is $\Delta T_b = K_b \times m$,where $m$ is the molality.
Given $m = 0.12 \ m$.
Therefore,$K_b = \frac{\Delta T_b}{m} = \frac{0.3 \ K}{0.12 \ mol \ kg^{-1}} = 2.5 \ K \ kg \ mol^{-1}$.
Thus,the correct option is $C$.

(C) The formula for boiling point elevation is $\Delta T_{b} = K_{b} \times m$,where $\Delta T_{b}$ is the elevation in boiling point,$K_{b}$ is the ebullioscopic constant,and $m$ is the molality of the solution.
Given: $\Delta T_{b} = 0.2 \ K$ and $K_{b} = 0.52 \ K \ kg \ mol^{-1}$.
Rearranging the formula to solve for molality $(m)$: $m = \frac{\Delta T_{b}}{K_{b}}$.
Substituting the values: $m = \frac{0.2}{0.52} \approx 0.3846 \ mol \ kg^{-1}$.
Rounding to three decimal places,we get $m \approx 0.385 \ mol \ kg^{-1}$.
Therefore,the correct option is $C$.

(A) The formula for boiling point elevation is $\Delta T_{b} = K_{b} \times m$,where $m$ is the molality of the solution.
Molality $(m)$ = $\frac{\text{moles of solute}}{\text{mass of solvent in kg}}$.
Moles of urea = $\frac{\text{given mass}}{\text{molar mass}} = \frac{15 \ g}{60 \ g \ mol^{-1}} = 0.25 \ mol$.
Mass of solvent (water) = $1000 \ g = 1 \ kg$.
Therefore,$m = \frac{0.25 \ mol}{1 \ kg} = 0.25 \ mol \ kg^{-1}$.
Now,$\Delta T_{b} = 0.52 \ K \ kg \ mol^{-1} \times 0.25 \ mol \ kg^{-1} = 0.13 \ K$.

(A) The elevation in boiling point is given by the formula: $\Delta T_{b} = K_{b} \times m$,where $m$ is the molality of the solution.
Molality $m = \frac{n_{solute}}{W_{solvent} \text{ (in kg)}}$.
Substituting the values: $0.3 \ K = 1.8 \ K \ kg \ mol^{-1} \times \frac{n_{solute}}{0.3 \ kg}$.
$n_{solute} = \frac{0.3 \times 0.3}{1.8} \ mol$.
$n_{solute} = \frac{0.09}{1.8} \ mol = 0.05 \ mol$.
Rounding to the nearest option,the correct value is $0.051 \ mol$.

(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 $m = \frac{\text{moles of solute}}{\text{mass of solvent in } kg}$.
Given: $\Delta T_{b} = 0.6 \ K$,$K_{b} = 2 \ K \ kg \ mol^{-1}$,and moles of solute $= 0.01 \ mol$.
Substituting the values: $0.6 = 2 \times \frac{0.01}{W_{solvent}}$.
$W_{solvent} = \frac{2 \times 0.01}{0.6} = \frac{0.02}{0.6} = \frac{1}{30} \ kg$.
$W_{solvent} \approx 0.0333 \ kg$.
Thus,the correct option is $C$.

(C) The formula for boiling point elevation is given by $\Delta T_b = K_b \times m$,where $\Delta T_b$ is the boiling point elevation,$K_b$ is the molal boiling point elevation constant,and $m$ is the molality of the solution.
Given: $\Delta T_b = 1.75 \ K$ and $K_b = 3 \ K \ kg \ mol^{-1}$.
Substituting the values into the formula: $1.75 = 3 \times m$.
Solving for $m$: $m = \frac{1.75}{3} \approx 0.5833 \ m$.
Rounding to two decimal places,we get $0.58 \ m$.

(C) The formula for boiling point elevation is given by: $\Delta T_{b} = K_{b} \times m$.
Here,$\Delta T_{b} = 0.39 \ K$ and $K_{b} = 0.52 \ K \ kg \ mol^{-1}$.
Rearranging the formula to solve for molality $(m)$: $m = \frac{\Delta T_{b}}{K_{b}}$.
Substituting the values: $m = \frac{0.39}{0.52} = 0.75 \ mol \ kg^{-1}$.
Therefore,the correct option is $C$.

(B) 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 $(m)$ is defined as the number of moles of solute $(n_{solute})$ divided by the mass of the solvent in kilograms $(W_{solvent(kg)})$.
So,$\Delta T_{b} = K_{b} \times \frac{n_{solute}}{W_{solvent(kg)}}$.
Given: $\Delta T_{b} = 0.8 \ K$,$K_{b} = 2 \ K \ kg \ mol^{-1}$,and $W_{solvent(kg)} = 0.5 \ kg$.
Substituting the values: $0.8 = 2 \times \frac{n_{solute}}{0.5}$.
$0.8 = 4 \times n_{solute}$.
$n_{solute} = \frac{0.8}{4} = 0.2 \ mol$.

(C) The elevation in boiling point is given by $\Delta T_{b} = T_{b} - T_{b}^{\circ} = 119.6^{\circ} C - 118^{\circ} C = 1.6 \ K$.
Using the formula $\Delta T_{b} = \frac{1000 \times K_{b} \times W_{2}}{M_{2} \times W_{1}}$,where $W_{2} = 5 \ g$,$W_{1} = 50 \ g$,and $K_{b} = 3.2 \ K \ kg \ mol^{-1}$.
Rearranging for molar mass $M_{2}$: $M_{2} = \frac{1000 \times 3.2 \times 5}{1.6 \times 50}$.
$M_{2} = \frac{16000}{80} = 200 \ g \ mol^{-1}$.

(C) The formula for boiling point elevation is $\Delta T_{b} = K_{b} \times m$.
Given $\Delta T_{b} = 1.75 \ K$ and $K_{b} = 3.5 \ K \ kg \ mol^{-1}$.
Substituting the values: $1.75 \ K = 3.5 \ K \ kg \ mol^{-1} \times m$.
Solving for molality: $m = \frac{1.75}{3.5} = 0.50 \ mol \ kg^{-1}$.

(A) The elevation in boiling point $(\Delta T_b)$ is given by the formula: $\Delta T_b = K_b \times m$,where $m$ is the molality of the solution.
Molality $(m)$ is defined as the number of moles of solute $(n_2)$ per kilogram of solvent $(W_1 \text{ in grams})$.
$m = \frac{n_2 \times 1000}{W_1} = \frac{W_2 \times 1000}{M_2 \times W_1}$.
Substituting this into the elevation formula: $\Delta T_b = \frac{K_b \times W_2 \times 1000}{M_2 \times W_1}$.
Rearranging for the molar mass of the solute $(M_2)$: $M_2 = \frac{1000 \times K_b \times W_2}{\Delta T_b \times W_1}$.

(B) The formula for elevation in boiling point is $\Delta T_{b} = K_{b} \times m$.
First,calculate the molality $(m)$: $m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.1 \ mol}{0.2 \ kg} = 0.5 \ mol \ kg^{-1}$.
Now,calculate the elevation in boiling point: $\Delta T_{b} = 0.513 \ ^{\circ}C \ kg \ mol^{-1} \times 0.5 \ mol \ kg^{-1} = 0.2565 \ ^{\circ}C$.
The boiling point of the solution $(T)$ is $T_{b} + \Delta T_{b} = 100 \ ^{\circ}C + 0.2565 \ ^{\circ}C = 100.2565 \ ^{\circ}C$,which is approximately $100.256 \ ^{\circ}C$.