Skip to content
Login Register

#Calculations

6 public questions tagged with this topic.

CHEMISTRY

Calculate the osmotic pressure of a solution containing 2.7 g of glucose ( C₆H₁₂O₆ ) in 250 mL of water at 27°C

Given: Calculate the osmotic pressure of a solution containing 2.7 g of glucose ( C₆H₁₂O₆ ) in 250 mL of water at 27°C. ( R = 0.0821 L atm/mol K, Molar mass of glucose = 180 g/mol ) These values define the system as per NCERT data. Formula: Moles of glucose = 2.7/180 = 0.015 mol. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Volume = 0.25 L. Molarity = 0.015/0.25 = 0.06 M . Pi = 0.06 × 0.0821 × 300 = 1.478 atm . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻Â

Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Some Basic Concepts, Structure of Atom and Periodicity, Topic: Mole concept, atomic models and periodic trends.

View Details
PHYSICS

Three charges +4 μC, -1 μC, and +2 μC are at (0, 0, 0), (4, 0, 0), and (0, 3, 0) m . What is the potential at (4, 3,

Given: Three charges +4 μC, -1 μC, and +2 μC are at (0, 0, 0), (4, 0, 0), and (0, 3, 0) m . What is the potential at (4, 3, 0) m ? (Take 1/4 π varepsilon_0 = 9 × 10⁹ Nm² C^{-2 ). These values define the system as per NCERT data. Formula: Distances: r_1 = sqrt4² + 3² = 5 m, r_2 = 3 m, r_3 = 4 m. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: V = 9 × 10⁹( frac4 × 10⁻⁶⁵+ frac-1 × 10⁻⁶³+ frac2 × 10⁻⁶⁴) . V = 9 × 10⁹( 0.8 × 10⁻⁶- 0.333 × 10⁻⁶+ 0.5 × 10⁻⁶) . V = 9 × 10⁹ × 0.967 × 10⁻⁶= 8703 V

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Electric Charges and Fields and Electrostatic Potential, Topic: Electric field, potential and capacitance concepts.

View Details