- A$K_2CrO_4$
- ✓$CrO_5$
- C$Cr_2(SO_4)_3$
- DNone
Step $(ii)$ $\left[ H _2 O _2 \rightarrow H _2 O +( O )\right] 4$
Step $(iii)$ $H _2 Cr _2 O _7+4( O ) \rightarrow 2 CrO _5+ H _2 O$
Hence, $K _2 Cr _2 O _7+ H _2 SO _4+4 H _2 O _2 \rightarrow 2 CrO _5+ K _2 SO _4+5 H _2 O$
The formation of Chromium Pentaoxide leads to the formation of blue colour from orange (as potassium dichromate is orange in colour). Chromium Pentaoxide is blue, so we get a blue colour after the reaction.
In the reaction, Hydrogen peroxide acts as an oxidizing agent, because it gets reduced, and its oxidation number changes to $-2$ from $-1$
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| Column $-I$ (various solutions) |
Column $-II$ (Their freezing point ) |
||
| $a$ | $0.1\,M$ $BaCl_2$ solution | $p$ | $271\,K$ |
| $b$ | $0.1\,M$ $NaCl$ solution | $q$ | $270\,K$ |
| $c$ | $0.1\,M\, K_3 [Fe(CN)_6]$ solution | $r$ | $268\,K$ |
| $d$ | $0.1\,M\, Al_2 (SO_4)_3$ solution | $s$ | $269\,K$ |
Given : Freezing point of $0.1\,M$ sucrose solution $= 272\,K$ and $F.pt.$ of water $= 273\,K$
Which of the following option show correct matches ?