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JEE Main 20252 AprilEvening Shift
Exam
JEE Main
Year
2025
Date
2 April
Shift
Evening Shift
Subject
Chemistry
Chapter
Solutions

JEE Main 2025 · 2 April · Evening Shift · Chemistry · Solutions

' x ' g of NaCl is added to water in a beaker with a lid. The...

' x ' g of NaCl is added to water in a beaker with a lid. The temperature of the system is raised from 1∘C1^{\circ} \mathrm{C} to 25∘C25^{\circ} \mathrm{C}. Which out of the following plots, is best suited for the change in the molarity (M) of the solution with respect to temperature?
[Consider the solubility of NaCl remains unchanged over the temperature range]' x ' g of NaCl is added to water in a beaker with a lid. The temperature of the system is raised from 1^ C to 25^ C .' x ' g of NaCl is added to water in a beaker with a lid. The temperature of the system is raised from 1^ C to 25^ C .' x ' g of NaCl is added to water in a beaker with a lid. The temperature of the system is raised from 1^ C to 25^ C .' x ' g of NaCl is added to water in a beaker with a lid. The temperature of the system is raised from 1^ C to 25^ C .

Options

  1. AThe foot of perpendicular of the point (2, 0, 5) on the line (x + 1)/(2) = (y - 1)/(5) = (z + 1)/(- 1) is (α, β, γ) .
  2. BThe foot of perpendicular of the point (2, 0, 5) on the line (x + 1)/(2) = (y - 1)/(5) = (z + 1)/(- 1) is (α, β, γ) .
  3. CThe foot of perpendicular of the point (2, 0, 5) on the line (x + 1)/(2) = (y - 1)/(5) = (z + 1)/(- 1) is (α, β, γ) .
  4. DThe foot of perpendicular of the point (2, 0, 5) on the line (x + 1)/(2) = (y - 1)/(5) = (z + 1)/(- 1) is (α, β, γ) .

Correct answer

B. The foot of perpendicular of the point (2, 0, 5) on the line (x + 1)/(2) = (y - 1)/(5) = (z + 1)/(- 1) is (α, β, γ) .

Step-by-step solution

 Water 1∘C→ decreases  Volume  Water 4∘C→ thermal expansion  Volume increases  Water 25∘C\underset{1^{\circ} \mathrm{C}}{\text{ Water }} \xrightarrow[\text{ decreases }]{\text{ Volume }}\underset{4^{\circ} \mathrm{C}}{\text{ Water }} \xrightarrow[\text{ thermal expansion }]{\text{ Volume increases }} \underset{25^{\circ} \mathrm{C}}{\text{ Water }} Water
 Molarity =nsolute ( Volume of solution )ℓ\text{ Molarity }=\frac{\mathrm{n}_{\text{solute }}}{(\text{ Volume of solution })_{\ell}}
Hence
1∘C→ increases  molarity 4∘C→ decreases  molarity 25∘1^{\circ} \mathrm{C} \xrightarrow[\text{ increases }]{\text{ molarity }} 4^{\circ} \mathrm{C} \xrightarrow[\text{ decreases }]{\text{ molarity }} 25^{\circ}C\mathrm{C}
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