- A$0.1$
- B$0.9$
- C$2$
- ✓$9$
$p H=2 ; H^{+}=10^{-2}=0.01\, M$
$\therefore M_{1}=0.1 ; V_{1}=1$
$M_{2}=0.01; V_{2}=?$
From
$M_{1} V_{1}=M_{2} V_{2} ; 0.1 \times 1=0.01 \times V_{2}$
$V_{2}=10$ litre
volume of water added $=10-1=9$ litre.
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$[$ Given $: \sqrt{3}=1.73, \sqrt{2}=1.41]$
(Consider heat capacity of all solutions as $4.2 J g ^{-1} K ^{-1}$ and density of all solutions as $1.0 \ g mL ^{-1}$ )
$1.$ Enthalpy of dissociation (in $kJ mol ^{-1}$ ) of acetic acid obtained from the Expt. $2$ is
$(A)$ $1.0$ $(B)$ $10.0$ $(C)$ $24.5$ $(D)$ $51.4$
$2.$ The $pH$ of the solution after Expt. $2$ is
$(A)$ $2.8$ $(B)$ $4.7$ $(C)$ $5.0$ $(D)$ $7.0$
Give the answer question $1$ and $2.$