1e:I[80107,["/_next/static/chunks/235ad55241b6443f.js","/_next/static/chunks/8337ab1f037b6654.js","/_next/static/chunks/c9e933b5fd1a0afc.js","/_next/static/chunks/848a330faa2461ac.js","/_next/static/chunks/9f2942e368a262d2.js","/_next/static/chunks/92f9e11244990470.js","/_next/static/chunks/a8b867fbb1dfca1a.js","/_next/static/chunks/1100f0390a901fa7.js","/_next/static/chunks/bbe6bd9d3c7b0355.js","/_next/static/chunks/a82195c3e5de7dd4.js"],"default"] 1f:I[76763,["/_next/static/chunks/235ad55241b6443f.js","/_next/static/chunks/8337ab1f037b6654.js","/_next/static/chunks/c9e933b5fd1a0afc.js","/_next/static/chunks/848a330faa2461ac.js","/_next/static/chunks/9f2942e368a262d2.js","/_next/static/chunks/92f9e11244990470.js","/_next/static/chunks/a8b867fbb1dfca1a.js","/_next/static/chunks/1100f0390a901fa7.js","/_next/static/chunks/bbe6bd9d3c7b0355.js","/_next/static/chunks/a82195c3e5de7dd4.js"],"default"] 1d:T439,{"@context":"https://schema.org","@graph":[{"@type":"BreadcrumbList","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https://www.divinejee.com"},{"@type":"ListItem","position":4,"name":"25 g of a nonvolatile nonelectrolyte is dissolved in 100 g o","item":"https://www.divinejee.com/question/25-g-of-a-nonvolatile-nonelectrolyte-is-dissolved-in-100-g-of-water-at-25-c-the-solution-showed"}]},{"@type":"QAPage","mainEntity":{"@type":"Question","name":"Question 25 g of a nonvolatile nonelectrolyte is dissolved in 100 g o on Topic","text":"

$2.5 \\mathrm{~g}$ of a non-volatile, non-electrolyte is dissolved in $100 \\mathrm{~g}$ of water at $25^{\\circ} \\mathrm{C}$. The solution showed a b...","answerCount":1,"upvoteCount":0,"datePublished":"2026-01-16","author":{"@type":"Organization","name":"DivineJEE"},"acceptedAnswer":{"@type":"Answer","text":"Check the detailed solution on the page.","upvoteCount":0,"url":"https://www.divinejee.com/question/25-g-of-a-nonvolatile-nonelectrolyte-is-dissolved-in-100-g-of-water-at-25-c-the-solution-showed"}}}]}20:T54e,

Molal boiling point elevation constant of water ($K_b$) = 0.52 K.kg.mol^-1
1 atm = 760 mm Hg
Molar mass of water = 18 g.mol^-1

First, we calculate the molality (m) of the solution using the boiling point elevation formula:

$$ \Delta T_b = K_b \times m $$

From the problem, we know:

$$ 2 = 0.52 \times m $$

Solving for m:

$$ m = \frac{2}{0.52} \approx 3.846 \text{ mol/kg} $$

Next, considering the solution is highly diluted, we use the formula for relative lowering of vapor pressure:

$$ \frac{\Delta P}{P^0} = \frac{n_{\text{solute}}}{n_{\text{solvent}}} $$

Given that molality (m) is defined as the number of moles of solute per kilogram of solvent:

$$ \frac{\Delta P}{P^0} = \frac{m}{1000} \times M_{solvent} $$

Substitute the known values:

$$ \Delta P = P^0 \times \frac{m}{1000} \times M_{solvent} $$

$$ = 760 \times \frac{3.846}{1000} \times 18 $$

$$ = 52.615 \text{ mm Hg} $$

Therefore, the vapor pressure of the solution:

$$ P_{solution} = P^0 - \Delta P $$

$$ = 760 - 52.615 $$

$$ \approx 707.385 \text{ mm Hg} $$

Rounding to the nearest integer, the vapor pressure of the aqueous solution is 707 mm Hg.