MCQ
How many protons are present in $1.8\,g \,NH_4^+$ ............. $N_A$
- A$1$
- B$1.2$
- ✓$1.1$
- D$11$
No. of $\mathrm{NH}_{4}^{+}$ ions in $1.8\, \mathrm{g}=\frac{1.8}{18} \times \mathrm{N}_{\mathrm{A}}=0.1\, \mathrm{N}_{\mathrm{A}}$
$\text { No. of protons in } 1.8\, \mathrm{g} \,\mathrm{NH}_{4}^{+} =11 \times 0.1\, \mathrm{N}_{\mathrm{A}}$
$=1.1 \,\mathrm{N}_{\mathrm{A}}$
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$(I)\,\,Ph-CH=CH-COOH$
$(II)\,\,p-NO_2-C_6H_4-CH=CH-COOH$
$(III)\,\,p-MeO-C_6H_4-CH=CH-COOH$
$(IV)\,\,p-Cl-C_6H_4-CH=CH-COOH$
(Plank's const. $ h = 6. \times 10^{-34}\, Js\,;$ mass of electron $= 9.1091 \times 10^{-31}\, kg\,;$ charge of electron $e= 1.60210 \times 10^{-19}\, C\,;$ permittivity of vaccum $\in _0 = 8.854185 \times 10^{-12} \,kg^{-1} \,m^{-3} A^2$)