MCQ
$23\, g$ of sodium will react with methyl alcohol to give
- Aone mole of oxygen
- B$22.4\, dm^3$ of hydrogen gas at $NTP$
- C$1\, mole$ of $H_2$
- ✓$1 1.2 \,L$ of hydrogen gas at $NTP$

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[Assume theat the pre-exponential factor is same in both the cases.
Given $R =8.31 \,J\, K ^{-1} \,mol ^{-1}$ ]

The above reaction is classified as

The limiting ionic conductivity ( $\Lambda 0$ ) values (in $mS m ^2 mol ^{-1}$ ) for different ions in aqueous solutions are given below:
| Ions | $Ag ^{+}$ | $K ^{+}$ | $Na ^{+}$ | $H ^{+}$ | $NO _3^{-}$ | $CI ^{-}$ | $SO _4^{2-}$ | $OH ^{-}$ | $CH_3COO ^{-}$ |
| $\Lambda_0$ | $6.2$ | $7.4$ | $5.0$ | $35.0$ | $7.2$ | $7.6$ | $16.0$ | $19.9$ | $4.1$ |
For different combinations of titrates and titrants given in $List-l$, the graphs of 'conductance' versus 'volume of titrant' are given in $List-II$.
Match each entry in $List-I$ with the appropriate entry in $List-II$ and choose the correct option.