MCQ
$1$ mol of $C{H_4}$contains
- A$6.02 \times {10^{23}}$atoms of $H$
- ✓$4 \,g$ atom of Hydrogen
- C$1.81 \times {10^{23}}$molecules of $C{H_4}$
- D$3.0 \,g$ of carbon
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(Given : $\left.pK _{ a }\left( CH _{3} COOH \right)=4.76\right)$
$\log 2=0.30$ $\log 3=0.48$ $\log 5=0.69$ $\log 7=0.84$ $\log 11=1.04$