A 1.0g sample of pyrolusite ore containing MnO2 (MW = 87) was dissolved in a concentrated HCl solution and liberated Cl2(g) was passed through a concentrated KI solution releasing I2. If the liberated iodine required 16mL of 1.25 M sodium thiosulphate (Na2S2O3) solution, mass percentage of MnO2 in the given sample is

No.of m.eqts. MnO2 = No.of m.eqts. of Cl2= No.of m.eqts. of KI = No.of m.eqts. hypoNo.of m.eqts of MnO2=16×1.25=16×54=20No.of milli moles of MnO2 = 10Weight of MnO2 = 10×87×10−3=0.87gWeight % of MnO2=0.871×100=87%

A 1.0g sample of pyrolusite ore containing MnO2 (MW = 87) was dissolved in a concentrated HCl solution and liberated Cl2(g) was passed through a concentrated KI solution releasing I2. If the liberated iodine required 16mL of 1.25 M sodium thiosulphate (Na2S2O3) solution, mass percentage of MnO2 in the given sample is

No.of m.eqts. MnO2 = No.of m.eqts. of Cl2= No.of m.eqts. of KI = No.of m.eqts. hypoNo.of m.eqts of MnO2=16×1.25=16×54=20No.of milli moles of MnO2 = 10Weight of MnO2 = 10×87×10−3=0.87gWeight % of MnO2=0.871×100=87%

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

A 1.0g sample of pyrolusite ore containing MnO₂ (MW = 87) was dissolved in a concentrated HCl solution and liberated Cl₂(g) was passed through a concentrated KI solution releasing I₂. If the liberated iodine required 16mL of 1.25 M sodium thiosulphate (Na₂S₂O₃) solution, mass percentage of MnO₂ in the given sample is

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.

An Intiative by Sri Chaitanya

43.5%

87%

21.75 %

0.875%

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

No.of m.eqts. MnO₂ = No.of m.eqts. of Cl₂
= No.of m.eqts. of KI = No.of m.eqts. hypo
No.of m.eqts of MnO2
$= 16 \times 1 .25 = 16 \times \frac{5}{4} = 20$
No.of milli moles of MnO₂ = 10
Weight of MnO₂ = $10 \times 87 \times 10^{- 3} = 0 .87 g$
Weight % of ${MnO}_{2} = \frac{0 .87}{1} \times 100 = 87 %$