H2O2 acts as both oxidising and reducing agent. As oxidising agent, its product is H2Obut as reducing agent, its product is O2. Volume strength has great significance for chemical reactions. The strength of ‘10V’ means 1 volume (or liter) of H2O2 on decomposition (H2O2→ H2O + 12 O2) gives 10 volumes (or liter) of oxygen at NTP.15 g Ba(MnO4)2 sample containing inert impurity is completely reacting with 100 mL of ‘11.2 V’ H2O2, then what will be the % purity of Ba(MnO4)2 in the sample : (Atomic mass : Ba = 137, Mn = 55)
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a
5%
b
10%
c
50%
d
none of these
answer is C.
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Detailed Solution
Ba(MnO4)2+5H2O2+3H2SO4→BaSO4+2MnSO4+5O2+8H2Omilliequivalents of Ba(MnO4)2 reacted = milliequivalents of H2O2 reactedmE×1000=NV(ml) ; EBa(MnO4)2 =37510 ; NH2O2 =vol.strength5.6 m375 × 10 × 1000=11.25.6 × 100∴ m=7.5 g∴ wt. of BaMnO42 in sample = 7.5 g% purity of BaMnO42 in sample = 7.515 × 100=50%
H2O2 acts as both oxidising and reducing agent. As oxidising agent, its product is H2Obut as reducing agent, its product is O2. Volume strength has great significance for chemical reactions. The strength of ‘10V’ means 1 volume (or liter) of H2O2 on decomposition (H2O2→ H2O + 12 O2) gives 10 volumes (or liter) of oxygen at NTP.15 g Ba(MnO4)2 sample containing inert impurity is completely reacting with 100 mL of ‘11.2 V’ H2O2, then what will be the % purity of Ba(MnO4)2 in the sample : (Atomic mass : Ba = 137, Mn = 55)