Q4 / 75 · JEE Main 2026 (05 April Shift 1) · Chemistry
What is the mole fraction of water in $10$% by weight (w/w) of aqueous urea solution?
[Given: Molar mass of H, O, C and N are $1$, $16$, $12$ and $14$ g mol$^{-1}$ respectively.]
[Given: Molar mass of H, O, C and N are $1$, $16$, $12$ and $14$ g mol$^{-1}$ respectively.]
- A$0.825$
- B$0.032$
- C$0.867$
- Correct answerD$0.967$
Solution
$10\%$ w/w aqueous urea solution means $10$ g of urea is present in $100$ g of solution.
Mass of urea $= 10$ g
Mass of water $= 100 - 10 = 90$ g
Molar mass of urea ($\text{NH}_2\text{CONH}_2$) $= 14 + 2 + 12 + 16 + 14 + 2 = 60$ g mol$^{-1}$
Molar mass of water ($\text{H}_2\text{O}$) $= 18$ g mol$^{-1}$
Moles of urea, $n_{\text{urea}} = \dfrac{10}{60} = \dfrac{1}{6}$ mol
Moles of water, $n_{\text{water}} = \dfrac{90}{18} = 5$ mol
Total moles $= 5 + \dfrac{1}{6} = \dfrac{31}{6}$ mol
Mole fraction of water, $x_{\text{water}} = \dfrac{n_{\text{water}}}{n_{\text{water}} + n_{\text{urea}}} = \dfrac{5}{\dfrac{31}{6}} = \dfrac{30}{31} \approx 0.967$
Answer: $0.967$
Mass of urea $= 10$ g
Mass of water $= 100 - 10 = 90$ g
Molar mass of urea ($\text{NH}_2\text{CONH}_2$) $= 14 + 2 + 12 + 16 + 14 + 2 = 60$ g mol$^{-1}$
Molar mass of water ($\text{H}_2\text{O}$) $= 18$ g mol$^{-1}$
Moles of urea, $n_{\text{urea}} = \dfrac{10}{60} = \dfrac{1}{6}$ mol
Moles of water, $n_{\text{water}} = \dfrac{90}{18} = 5$ mol
Total moles $= 5 + \dfrac{1}{6} = \dfrac{31}{6}$ mol
Mole fraction of water, $x_{\text{water}} = \dfrac{n_{\text{water}}}{n_{\text{water}} + n_{\text{urea}}} = \dfrac{5}{\dfrac{31}{6}} = \dfrac{30}{31} \approx 0.967$
Answer: $0.967$
Hence, the correct answer is option: 'D'.