MCQ
How many gm of bromine will react with $21\, gm$ ${C_3}{H_6}$
- ✓$80$
- B$160$
- C$240$
- D$320$
$42\, gm$ of propene reacts with $160\, gms$ of bromine.
$\therefore $ $21\,gms$ of propene $\frac{{160}}{{42}} \times \,21 = 80\,gms$.
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$[\log 5=0.698, \log 7=0.845, \log 2=0.301]$
$(i) C_2H_5OH \longrightarrow CH_3CHO$
$(ii) C_2H_5OH \longrightarrow C_2H_5ONa$
$CH_2OH-CHOH-CH_2OH$ $\xrightarrow{{KHS{O_4}/\Delta }}(X)\mathop {\xrightarrow{{{{({C_2}{H_5}O)}_3}Al}}}\limits_\Delta (Y)$
$(Y)$ will be: