Question
An element $'P'$ has atomic number $56.$ What will be the formula of its halide ?
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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 ?
$Pt\,|\,B{r_{2(\ell )0.1M}}\,|\,Br_{({\text{aq}}),0.1M}^ - \,||\,H_{_{({\text{aq}})0.1M}}^ + \,|\,{H_{2(g)1atm}}\,|\,Pt$ given ; $E_{B{r^ - }/B{r_2}}^0\, = \, - 1.06\,V$