Elevation of boiling point of the solvent Questions in English

(C) The elevation in boiling point $(\Delta T_b)$ is a colligative property,which depends on the van't Hoff factor $(i)$ and the molality $(m)$ of the solution: $\Delta T_b = i \times K_b \times m$.
Upon complete dissociation,the number of particles produced per formula unit is:
$(A)$ $0.1 \ m \ KCl \rightarrow i = 2$,particles = $0.1 \times 2 = 0.2 \ m$
$(B)$ $0.05 \ m \ NaCl \rightarrow i = 2$,particles = $0.05 \times 2 = 0.1 \ m$
$(C)$ $0.1 \ m \ BaCl_2 \rightarrow i = 3$,particles = $0.1 \times 3 = 0.3 \ m$
$(D)$ $0.1 \ m \ MgSO_4 \rightarrow i = 2$,particles = $0.1 \times 2 = 0.2 \ m$
Since the $0.1 \ m \ BaCl_2$ solution has the highest concentration of particles $(0.3 \ m)$,it exhibits the maximum elevation in boiling point.

(B) The formula for boiling point elevation is $\Delta T_{b} = K_{b} \times m$.
Given: $\Delta T_{b} = 0.5 \ K$ and $K_{b} = 2.40 \ K \ kg \ mol^{-1}$.
Substituting the values: $0.5 \ K = 2.40 \ K \ kg \ mol^{-1} \times m$.
Therefore,$m = \frac{0.5}{2.40} \approx 0.2083 \ mol \ kg^{-1}$.
Rounding to two decimal places,we get $m = 0.21 \ mol \ kg^{-1}$.

(A) Given:
- For $0.1 \ molal$ solution,$T_b = 100.16^{\circ} C$.
- Boiling point of pure water is $100^{\circ} C$.
- Boiling point elevation formula: $\Delta T_b = K_b \cdot m$ (since glucose is a non-electrolyte,$i = 1$).
Step $1$: Calculate elevation for $0.1 \ molal$ solution:
$\Delta T_b = 100.16 - 100 = 0.16^{\circ} C$.
Step $2$: Use the proportionality $\Delta T_b \propto m$ to find elevation for $0.5 \ molal$ solution:
$\Delta T_b^{\prime} = \Delta T_b \times \frac{m^{\prime}}{m} = 0.16 \times \frac{0.5}{0.1} = 0.16 \times 5 = 0.8^{\circ} C$.
Step $3$: Calculate new boiling point:
$T_b^{\prime} = 100 + 0.8 = 100.80^{\circ} C$.
Thus,the correct option is $A$.

(B) The formula for boiling point elevation is $\Delta T_{b} = K_{b} \cdot m$,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.
Substituting the values: $\Delta T_{b} = 0.01 \ K$,$K_{b} = 0.50 \ K \ kg \ mol^{-1}$,$W_2 = 0.35 \ g$,and $W_1 = 100 \ g$.
$0.01 = 0.50 \times \frac{0.35 \times 1000}{M_2 \times 100}$.
$M_2 = \frac{0.50 \times 0.35 \times 1000}{0.01 \times 100} = \frac{175}{1} = 175 \ g \ mol^{-1}$.

(C) The formula for molar mass of a solute is given by: $M_2 = \frac{K_b \times W_2 \times 1000}{\Delta T_b \times W_1}$
Given: $K_b = 2.5 \ K \ kg \ mol^{-1}$,$W_2 = 3.5 \ g$,$W_1 = 100 \ g$,and $\Delta T_b = 0.35 \ K$.
Substituting the values: $M_2 = \frac{2.5 \times 3.5 \times 1000}{0.35 \times 100}$
$M_2 = \frac{8750}{35} = 250 \ g \ mol^{-1}$.

(B) Boiling point elevation is a colligative property given by $\Delta T_b = i \cdot K_b \cdot m$. Assuming complete dissociation,we compare the product $i \cdot m$ for each solution:
$(A)$ For $0.1 \text{ m } AlCl_3$: $i = 4$,$i \cdot m = 4 \times 0.1 = 0.4$
$(B)$ For $0.01 \text{ m } MgCl_2$: $i = 3$,$i \cdot m = 3 \times 0.01 = 0.03$
$(C)$ For $1 \text{ m } KCl$: $i = 2$,$i \cdot m = 2 \times 1 = 2.0$
$(D)$ For $0.5 \text{ m } NaCl$: $i = 2$,$i \cdot m = 2 \times 0.5 = 1.0$
The lowest value of $i \cdot m$ is $0.03$,hence $0.01 \text{ m } MgCl_2$ has the lowest boiling point elevation.

(D) The molality $m$ of the solution is calculated as follows:
Mass of glucose $(W_2)$ $= 36 \ g$.
Molar mass of glucose $(M_2)$ $= 180 \ g \ mol^{-1}$.
Volume of solution $= 1 \ dm^3 = 1 \ L$.
Assuming the density of the solution is approximately $1 \ g \ mL^{-1}$,the mass of the solvent (water) $W_1$ is $1000 \ g = 1 \ kg$.
Molality $m = \frac{W_2}{M_2 \times W_1 (\text{in } kg)} = \frac{36}{180 \times 1} = 0.2 \ mol \ kg^{-1} = \frac{2}{10} \ mol \ kg^{-1}$.
The elevation in boiling point is given by $\Delta T_b = K_b \times m = K_b \times \frac{2}{10} = \frac{2 \ K_b}{10}$.
The boiling point of water is $100^{\circ} C$.
Therefore,the boiling point of the solution $T_b = 100 + \Delta T_b = (100 + \frac{2 \ K_b}{10})^{\circ} C$.

(A) The formula for boiling point elevation 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 $W_2 = 1.5 \ g$,$M_2 = 150 \ g \ mol^{-1}$,and $W_1 = 30 \ g$.
Substituting these into the equation $\Delta T_b = K_b \times \frac{W_2 \times 1000}{M_2 \times W_1}$:
$0.65 = K_b \times \frac{1.5 \times 1000}{150 \times 30}$
$0.65 = K_b \times \frac{1500}{4500}$
$0.65 = K_b \times \frac{1}{3}$
$K_b = 0.65 \times 3 = 1.95 \ K \ kg \ mol^{-1}$.

(C) The formula for boiling point elevation is $\Delta T_{b} = K_{b} \times m$.
Given: $\Delta T_{b} = 1.89 \ K$ and $K_{b} = 3.15 \ K \ kg \ mol^{-1}$.
Substituting the values: $1.89 = 3.15 \times m$.
Therefore,$m = \frac{1.89}{3.15} = 0.6 \ mol \ kg^{-1}$ or $0.6 \ m$.

(D) The boiling point elevation $\Delta T_b$ is given by $\Delta T_b = i \times K_b \times m$,where $i$ is the van't Hoff factor and $m$ is the molality. The value of $i \times m$ determines the magnitude of elevation.
$A) \ 0.1 \ m \ NaCl: i \times m = 2 \times 0.1 = 0.2 \ m$
$B) \ 0.2 \ m \ KNO_3: i \times m = 2 \times 0.2 = 0.4 \ m$
$C) \ 0.1 \ m \ Na_2SO_4: i \times m = 3 \times 0.1 = 0.3 \ m$
$D) \ 0.05 \ m \ CaCl_2: i \times m = 3 \times 0.05 = 0.15 \ m$
Since $0.05 \ m \ CaCl_2$ has the lowest value of $i \times m$ $(0.15)$,it exhibits the minimum boiling point elevation.

(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 $(m)$ is defined as the number of moles of solute per kilogram of solvent.
Given that $1 \ \text{mole}$ of solute is dissolved in $1 \ kg$ of solvent,the molality $m = \frac{1 \ \text{mol}}{1 \ \text{kg}} = 1 \ \text{mol} \ kg^{-1}$.
Substituting the values into the equation: $x \ K = K_{b} \times 1 \ \text{mol} \ kg^{-1}$.
Therefore,$K_{b} = x \ K \ kg \ mol^{-1}$.

(A) 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.
Substituting the values: $1.75 = \frac{3 \times 5.6 \times 1000}{M_2 \times 50}$.
Rearranging for $M_2$: $M_2 = \frac{3 \times 5.6 \times 1000}{1.75 \times 50}$.
$M_2 = \frac{16800}{87.5} = 192 \ g \ mol^{-1}$.

(C) The formula for boiling point elevation is $\Delta T_{b} = K_{b} \times m$.
$\therefore m = \frac{\Delta T_{b}}{K_{b}}$.
Substituting the given values: $m = \frac{7.15 \ K}{2.75 \ K \ kg \ mol^{-1}}$.
$m = 2.6 \ mol \ kg^{-1}$ (or $2.6 \ m$).

(A) The boiling point elevation 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 ($W_1$ in grams): $m = \frac{n_2 \times 1000}{W_1}$.
Since $n_2 = \frac{W_2}{M_2}$,where $W_2$ is the mass of solute and $M_2$ is the molar mass of solute,we substitute this into the molality equation:
$m = \frac{W_2 \times 1000}{M_2 \times W_1}$.
Substituting this into the boiling point elevation formula:
$\Delta T_b = K_b \times \frac{W_2 \times 1000}{M_2 \times W_1}$.
Rearranging for $M_2$,we get: $M_2 = \frac{1000 \times K_b \times W_2}{\Delta T_b \times W_1}$.

(A) The elevation in boiling point is given by $\Delta T_{b} = T_{b} - T_{b}^{\circ} = 85^{\circ}C - 76^{\circ}C = 9^{\circ}C$.
Using the formula $\Delta T_{b} = K_{b} \times m$,where $m$ is the molality:
$9 = 2.7 \times \frac{W_{B}}{120} \times \frac{1000}{160}$.
Solving for $W_{B}$:
$W_{B} = \frac{9 \times 120 \times 160}{2.7 \times 1000} = \frac{172800}{2700} = 64 \ g$.

(C) The boiling point of a liquid is defined as the temperature at which its vapour pressure becomes equal to the applied or external pressure.

(A) Given: $W_2 = 6 \ g$ (mass of solute),$W_1 = 100 \ g$ (mass of solvent),$K_b = 0.52 \ K \ kg \ mol^{-1}$,$T_b = 100.52^{\circ} C$.
Elevation in boiling point $\Delta T_b = T_b - T_b^{\circ} = 100.52^{\circ} C - 100^{\circ} C = 0.52 \ K$.
Using the formula for molar mass of solute: $M_2 = \frac{1000 \cdot K_b \cdot W_2}{\Delta T_b \cdot W_1}$.
Substituting the values: $M_2 = \frac{1000 \times 0.52 \times 6}{0.52 \times 100}$.
$M_2 = \frac{1000 \times 6}{100} = 60 \ g \ mol^{-1}$.

(D) The formula for elevation in boiling point is $\Delta T_b = K_b \times m$.
Given that the solution is decimolal,the molality $m = 0.1 \ mol \ kg^{-1}$.
Given $K_b = 0.52 \ ^{\circ}C \ kg \ mol^{-1}$.
Calculating the elevation in boiling point: $\Delta T_b = 0.52 \times 0.1 = 0.052 \ ^{\circ}C$.
The boiling point of pure water $T_b^{\circ}$ is $100 \ ^{\circ}C$.
The boiling point of the solution $T_b$ is given by $T_b = T_b^{\circ} + \Delta T_b$.
$T_b = 100 \ ^{\circ}C + 0.052 \ ^{\circ}C = 100.052 \ ^{\circ}C$.

(B) The elevation in boiling point is given by $\Delta T_{b} = K_{b} \cdot m$.
For the glucose solution: $T_{b} - T_{b}^{\circ} = K_{b} \cdot m$.
$100.16^{\circ} C - 100^{\circ} C = K_{b} \times 0.1 \ m$.
$0.16 = K_{b} \times 0.1$,so $K_{b} = 1.6 \ ^{\circ} C \ kg \ mol^{-1}$.
Now,for the sucrose solution: $\Delta T_{b} = K_{b} \times m = 1.6 \times 0.5 = 0.80^{\circ} C$.
The boiling point of the sucrose solution is $T_{b} = T_{b}^{\circ} + \Delta T_{b} = 100^{\circ} C + 0.80^{\circ} C = 100.80^{\circ} C$.

(D) The elevation in boiling point is given by $\Delta T_{b} = K_{b} \cdot m$.
Since both solutions boil at the same temperature,their boiling point elevation $\Delta T_{b}$ must be equal.
Assuming $K_{b}$ is the same for both dilute aqueous solutions,the molality $m$ must be equal.
Molality $m = \frac{\text{moles of solute}}{\text{mass of solvent in } kg}$.
For $1 \ dm^3$ of water,the mass is $1 \ kg$.
For urea: $\text{moles} = \frac{3 \ g}{60 \ g \ mol^{-1}} = 0.05 \ mol$.
For solute $A$: $\text{moles} = \frac{4.5 \ g}{M_A}$.
Equating the molalities: $0.05 = \frac{4.5}{M_A}$.
$M_A = \frac{4.5}{0.05} = 90 \ g \ mol^{-1}$.

(D) The elevation in boiling point is given by $\Delta T_b = T_b - T_b^{\circ} = 100.18^{\circ}C - 100^{\circ}C = 0.18 \ K$.
Given $K_b = 0.512 \ K \ kg \ mol^{-1}$.
Using the formula $\Delta T_b = K_b \times m$,where $m$ is the molality.
$m = \frac{\Delta T_b}{K_b} = \frac{0.18}{0.512} \approx 0.35 \ mol \ kg^{-1}$.

(D) The elevation in boiling point is given by the formula: $\Delta T_{b} = K_{b} \times m$.
Here,$m$ is the molality,which is defined as $m = \frac{W_{2} \times 1000}{M_{2} \times W_{1}}$.
Substituting this into the equation: $\Delta T_{b} = K_{b} \times \frac{W_{2} \times 1000}{M_{2} \times W_{1}}$.
Rearranging for $K_{b}$: $K_{b} = \frac{\Delta T_{b} \times M_{2} \times W_{1}}{W_{2} \times 1000}$.
Note: If the molality is defined as moles per kg of solvent,the $1000$ factor is omitted. Given the options provided,option $D$ represents the relationship where $W_{1}$ is in kg.

(D) The molal elevation constant $(K_b)$ is defined as the elevation in boiling point when $1 \ mole$ of a non-volatile solute is dissolved in $1 \ kg$ $(1000 \ g)$ of solvent.
This is based on the formula $\Delta T_b = K_b \times m$,where $m$ is the molality of the solution.
When $m = 1 \ mol/kg$,then $\Delta T_b = K_b$.

(C) Key Idea: The elevation in boiling point is directly proportional to the molal concentration of the solute in a solution,given by the formula: $\Delta T_b = K_b \times m$.
Given:
Molality $(m)$ = $0.25 \ mol \ kg^{-1}$
Ebullioscopic constant $(K_b)$ = $0.52 \ K \ kg \ mol^{-1}$
Calculation:
$\Delta T_b = 0.52 \times 0.25 = 0.13 \ K$
Therefore,the elevation in boiling point is $0.13 \ K$.

(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 $m$ is defined as the number of moles of solute $(n_2)$ per kilogram of solvent ($W_1$ in grams): $m = \frac{W_2 \times 1000}{M_2 \times W_1}$.
Substituting this into the elevation formula: $\Delta T_b = \frac{K_b \times W_2 \times 1000}{M_2 \times W_1}$.
Rearranging to solve for the molar mass of the solute $(M_2)$: $M_2 = \frac{K_b \times W_2 \times 1000}{\Delta T_b \times W_1}$.

(A) The formula for elevation in boiling point is given by $\Delta T_b = K_b \times m$,where $\Delta T_b$ is the elevation in boiling point,$K_b$ is the molal elevation constant,and $m$ is the molality of the solution.
Given values are $\Delta T_b = 0.16 \ K$ and $m = 0.05 \ m$.
Substituting these values into the formula: $0.16 = K_b \times 0.05$.
Solving for $K_b$: $K_b = \frac{0.16}{0.05} = 3.2 \ K \ kg \ mol^{-1}$.
Therefore,the correct option is $A$.

(D) The boiling point elevation is given by $\Delta T_b = T_b - T_b^{\circ} = 102.2^{\circ} C - 100^{\circ} C = 2.2 \ K$.
Using the formula $\Delta T_b = i \times K_b \times m$,where $K_b = 0.512 \ K \ kg \ mol^{-1}$ and $m = 1 \ m$.
For $1 \ m \ NaCl$,the van't Hoff factor $i \approx 2$.
$\Delta T_b = 2 \times 0.512 \times 1 = 1.024 \ K$.
If we assume the question implies a specific $K_b$ value of $2.2 \ K \ kg \ mol^{-1}$ as stated in the prompt,then for $1 \ m$ non-electrolyte (glucose),$\Delta T_b = 1 \times 2.2 \times 1 = 2.2 \ K$.
Thus,$T_b = 100 + 2.2 = 102.2^{\circ} C$.
Therefore,$1 \ m \ \text{glucose}$ is the correct answer based on the provided constant.

(D) The elevation in boiling point is given by the formula: $\Delta T_{b} = i \times K_{b} \times m$.
For glucose,which is a non-electrolyte,the van't Hoff factor $(i)$ is $1$.
Given the molality $(m)$ is $1 \ m$,we substitute these values into the equation:
$\Delta T_{b} = 1 \times K_{b} \times 1 = K_{b}$.
Therefore,$\Delta T_{b} = K_{b}$.

(B) 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 $m = 1 \ m$,we have $\Delta T_b = 0.52 \times 1 = 0.52 \ K$.
The boiling point of pure water is $373.15 \ K$.
Therefore,the boiling point of the solution is $T_b = T_b^0 + \Delta T_b = 373.15 \ K + 0.52 \ K = 373.67 \ K$.
Thus,the correct option is $B$.

(A) The elevation in boiling point is given by the formula: $\Delta T_b = i \times K_b \times m$.
Since glucose is a non-electrolyte,the van't Hoff factor $i = 1$.
The molality $m$ is calculated as: $m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{180 \ g / 180 \ g \ mol^{-1}}{1 \ kg} = 1 \ mol \ kg^{-1}$.
Substituting the values: $\Delta T_b = 1 \times 0.52 \ K \ kg \ mol^{-1} \times 1 \ mol \ kg^{-1} = 0.52 \ K$.
The boiling point of the solution $T_b$ is: $T_b = T_b^0 + \Delta T_b = 373.15 \ K + 0.52 \ K = 373.67 \ K$.

(A) Given: Mass of glucose $(w) = 1.8 \ g$
Mass of solvent $(W) = 100 \ g$
Elevation in boiling point $(\Delta T_b) = 0.1^{\circ} C$
Molar mass of glucose $(M) = 180 \ g / mol$
The formula for elevation in boiling point is $\Delta T_b = K_b \times m$,where $m$ is molality.
Molality $(m) = \frac{w \times 1000}{M \times W} = \frac{1.8 \times 1000}{180 \times 100} = 0.1 \ mol / kg$
Now,$K_b = \frac{\Delta T_b}{m} = \frac{0.1}{0.1} = 1 \ K \ kg / mol$.

(C) Boiling occurs at the temperature where the vapour pressure of the liquid becomes equal to the external atmospheric pressure.
According to the graph,the upper curve represents pure water and the lower curve represents the aqueous urea solution (since the addition of a non-volatile solute lowers the vapour pressure).
The boiling point of the urea solution is the temperature at which its vapour pressure reaches $1.0 \ atm$.
Looking at the graph,the lower curve intersects the $1.0 \ atm$ line at temperature $T_3$.

(B) The boiling point elevation is given by the formula $\Delta T_{b} = i \times K_{b} \times m$.
For $KCl$,the van't Hoff factor $i = 2$ because it dissociates completely $(\alpha = 100 \ \%)$ into $K^+$ and $Cl^-$.
Given $K_{b} = 0.52 \ K \ kg \ mol^{-1}$ and $m = 0.1 \ m$.
$\Delta T_{b} = 2 \times 0.52 \times 0.1 = 0.104 \ K$.
The boiling point of the solution $T_{b} = T_{b}^{0} + \Delta T_{b} = 373 \ K + 0.104 \ K = 373.104 \ K$.

(B) The elevation in boiling point $(\Delta T_{b})$ is directly proportional to the molality $(m)$ of the solution when a non-volatile solute is added to a volatile solvent.
$\Delta T_{b} = K_{b} \times m$
Where:
$\Delta T_{b}$ is the elevation in boiling point.
$K_{b}$ is the molal elevation constant (ebullioscopic constant).
$m$ is the molality of the solution.
Rearranging the formula:
$K_{b} = \frac{\Delta T_{b}}{m}$
Therefore,the molal elevation constant is the ratio of the elevation in boiling point to the molality.

(B) Mass of solute $X = 9 \ g$,Mass of solute $Y = 1 \ g$.
Mass of solvent $= 100 - (9 + 1) = 90 \ g$.
Moles of $X = \frac{9}{180} = 0.05 \ mol$.
Moles of $Y = \frac{1}{50} = 0.02 \ mol$.
Total moles of solute $= 0.05 + 0.02 = 0.07 \ mol$.
Molality $(m) = \frac{\text{Total moles of solute}}{\text{Mass of solvent in kg}} = \frac{0.07}{0.090} \approx 0.777 \ mol \ kg^{-1}$.
Elevation in boiling point $\Delta T_b = K_b \times m = 0.52 \times 0.777 \approx 0.404 \ { }^{\circ} C$.
Boiling point of solution $= 100 + 0.404 = 100.404 \ { }^{\circ} C \approx 100.4 \ { }^{\circ} C$.
Thus,option $(B)$ is the correct answer.

(A) The elevation in boiling point is given by $\Delta T_b = T_b - T_b^{\circ} = K_b \times m$,where $m$ is the molality of the solution.
$\Delta T_b = 373.202 \ K - 373.15 \ K = 0.052 \ K$.
Using the formula $\Delta T_b = K_b \times \frac{x \times 1000}{y \times M_{\text{urea}}}$,where $M_{\text{urea}} = 60 \ g \ mol^{-1}$.
$0.052 = 0.52 \times \frac{x \times 1000}{y \times 60}$.
$0.052 = \frac{x}{y} \times \frac{520}{60}$.
$\frac{x}{y} = \frac{0.052 \times 60}{520} = \frac{3.12}{520} = 0.006 = 6.0 \times 10^{-3}$.

(B) This question is based on the colligative properties of solutions.
Colligative properties are those properties of a solution that depend only on the number of solute particles and not on the nature of the solute.
When a non-volatile solute like salt $(NaCl)$ is added to 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 than the pure solvent.
Therefore,the boiling point of the solution is higher than the boiling point of the pure solvent,a phenomenon known as elevation in boiling point.

(B) The formula for boiling point elevation is $\Delta T_b = K_b \times m$.
Given,$K_b = 0.52 \ K \ kg \ mol^{-1}$ and molality $m = 0.1 \ m$.
Calculating the elevation in boiling point: $\Delta T_b = 0.52 \times 0.1 = 0.052 \ K$ (or $^{\circ}C$).
The boiling point of the solution is $T_b = T_b^{\circ} + \Delta T_b$.
Since the boiling point of pure water $T_b^{\circ} = 100^{\circ}C$,we have $T_b = 100 + 0.052 = 100.052^{\circ}C$.

(D) The elevation in boiling point is given by the formula: $\Delta T_b = K_b \cdot m$.
Here,$\Delta T_b = T_b - T_b^{\circ}$.
Given: $T_b = 100.52^{\circ} C$,$T_b^{\circ} = 100^{\circ} C$,and $K_b = 0.52 \ K \ kg \ mol^{-1}$.
Calculating the elevation: $\Delta T_b = 100.52 - 100 = 0.52 \ K$.
Substituting the values into the formula: $0.52 = 0.52 \times m$.
Therefore,$m = \frac{0.52}{0.52} = 1.0 \ m$.

(C) The formula for elevation of boiling point is $\Delta T_{b} = \frac{w \times K_{b} \times 1000}{M \times W}$.
Here,$w$ is the weight of the solute,$W$ is the weight of the solvent $(100 \ g)$,$M$ is the molecular weight of the solute $(100)$,and $K_{b}$ is the ebullioscopic constant $(0.5 \ K \ kg \ mol^{-1})$.
Given $\Delta T_{b} = 0.05^{\circ} C$.
Substituting the values into the formula:
$0.05 = \frac{w \times 0.5 \times 1000}{100 \times 100}$.
$0.05 = \frac{w \times 500}{10000} = \frac{w}{20}$.
$w = 0.05 \times 20 = 1 \ g$.

(A) Elevation in boiling point is given by $\Delta T_{b} = i \times K_{b} \times m$.
For $NaCl$ $(58.4 \ g \approx 1 \text{ mole})$,$i = 2$.
For glucose $(180 \ g = 1 \text{ mole})$,$i = 1$.
Since both solutes are dissolved in the same amount of solvent ($1000 \ mL \approx 1 \text{ kg}$ of water),their molality $(m)$ is the same.
Therefore,$\Delta T_{b} \propto i$.
Since $i_{NaCl} > i_{glucose}$,the $NaCl$ solution will show a higher elevation of boiling point.