0.02 g of the following substances (assuming non-electrolytes) are dissolved in 500 mL of water. Arrange them in increasing order of their elevation of boiling point of solution.
Given: Molar mass of H, C, N, O and Cl are 1, 12, 14, 16 and 35.5 g mol-1 respectively.
(A) C6H12O6
(B) NH2CONH2
(C) C12H22O11
(D) C3H7Cl
Choose the correct order from the options given below.
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (C), (A), (D), (B)
Elevation of boiling point is a colligative property: it depends on the number of solute particles, not on their nature. For a non-electrolyte:
\( \displaystyle \Delta T_b=K_b\,m=K_b\times\frac{w_2\times1000}{M_2\times w_1} \)
Here w2 and M2 are the mass and molar mass of the solute, and w1 is the mass of the solvent in grams.
In this question, the same mass of each solute (0.02 g) is dissolved in the same amount of water. So ΔTb is proportional to 1/M2: the smaller the molar mass, the more particles, and the larger the elevation.
First, (C) sucrose: \( M = 12(12)+22(1)+11(16) = 342\ \mathrm{g\,mol^{-1}} \). The largest molar mass gives the fewest particles and the smallest elevation.
Second, (A) glucose: \( M = 6(12)+12(1)+6(16) = 180\ \mathrm{g\,mol^{-1}} \).
Third, (D): \( \ce{C3H7Cl} \), \( M = 3(12)+7(1)+35.5 = 78.5\ \mathrm{g\,mol^{-1}} \).
Fourth, (B) urea: \( M = 2(14)+4(1)+12+16 = 60\ \mathrm{g\,mol^{-1}} \). The smallest molar mass gives the most particles and the largest elevation.
Shortcut: There is no need to calculate ΔTb. Just arrange the molar masses from largest to smallest: sucrose (C) first and urea (B) last. Only Option (2) does this.