The students of Class XII were on a study trip. They visited a nearby lake and took some water…
The students of Class XII were on a study trip. They visited a nearby lake and took some water samples which were rich in sodium chloride salt. The boiling point of lake water was found to be $373\cdot032\text{ K}$.
If $500\text{ g}$ of lake water sample was used which contained $0\cdot45\text{ g}$ of NaCl, what is the observed molecular weight of NaCl in the lake sample? Assume that NaCl ionises completely in the water sample. ($K_b$ for water $= 0\cdot52\text{ K kg mol}^{-1}$)
Answer
Answer
AIWritten by AI (gemini) - it can contain mistakes.
Given: Boiling point $= 373\cdot032\text{ K}$, $\Delta T_b = 373\cdot032 - 373\cdot000 = 0\cdot032\text{ K}$.
Mass of solute $w = 0\cdot45\text{ g}$, mass of solvent $W = 500\text{ g}$, $K_b = 0\cdot52\text{ K kg mol}^{-1}$.
For complete ionisation of $\text{NaCl} \rightarrow \text{Na}^+ + \text{Cl}^-$, $i = 2$.
$\Delta T_b = \frac{i \times K_b \times w \times 1000}{M_{\text{obs}} \times W}$
$M_{\text{obs}} = \frac{2 \times 0\cdot52 \times 0\cdot45 \times 1000}{0\cdot032 \times 500} = \frac{468}{16} = 29\cdot25\text{ g mol}^{-1}$
Final answer: 29.25 g mol-1
Final answer: 29.25 g mol-1
From ISC 2027 Specimen Chemistry Paper 1, question 11.