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NEET CHEMISTRYSolutionsEasy

Question

Normality of a solution containing 9.8 g9.8 \text{ g} of H2SO4\text{H}_2\text{SO}_4 in 250 cm3250 \text{ cm}^3 of the solution is

A

0.8 N0.8 \text{ N}

B

1 N1 \text{ N}

C

0.08 N0.08 \text{ N}

D

1.8 N1.8 \text{ N}

Step-by-Step Solution

Given: Mass of H2SO4\text{H}_2\text{SO}_4 (ww) = 9.8 g9.8 \text{ g} Volume of solution (VV) = 250 cm3=250 mL=0.25 L250 \text{ cm}^3 = 250 \text{ mL} = 0.25 \text{ L}

Molar mass of H2SO4\text{H}_2\text{SO}_4 (MM) = 2(1)+32+4(16)=98 g/mol2(1) + 32 + 4(16) = 98 \text{ g/mol} Since H2SO4\text{H}_2\text{SO}_4 is a dibasic acid, its n-factor = 22. Equivalent mass of H2SO4\text{H}_2\text{SO}_4 (EE) = Molar massn-factor=982=49 g/eq\frac{\text{Molar mass}}{\text{n-factor}} = \frac{98}{2} = 49 \text{ g/eq}

Number of gram equivalents of H2SO4\text{H}_2\text{SO}_4 = MassEquivalent mass=9.849=0.2 eq\frac{\text{Mass}}{\text{Equivalent mass}} = \frac{9.8}{49} = 0.2 \text{ eq}

Normality (NN) = Number of gram equivalents of soluteVolume of solution in L\frac{\text{Number of gram equivalents of solute}}{\text{Volume of solution in L}} N=0.2 eq0.25 L=0.8 NN = \frac{0.2 \text{ eq}}{0.25 \text{ L}} = 0.8 \text{ N}

Exam Context & Concepts Covered

This question aligns with the NEET CHEMISTRY syllabus, specifically targeting concepts from Solutions. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

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