MCQ
In the preceeding question, the amount of $N{a_2}C{O_3}$ present in the solution is ............. $\mathrm{g}$
- A$2.650$
- B$1.060$
- ✓$0.530$
- D$0.265$
Meq. of $N{a_2}C{O_3}$+ Meq. of $NaOH$ = Meq. of $HCl$
$\frac{W}{E} \times 1000 + \frac{W}{E} \times 1000 = NV$
(Suppose $N{a_2}C{O_3} = a\,gm$, $NaOH = b\, gm$)
$\frac{a}{{106}} \times 1000 + \frac{b}{{40}} \times 1000 = 300 \times 0.1$.....$(1)$
From solution of $(31)$ From equation $(1)$ $a = N{a_2}C{O_3} = 0.53\,gm$.
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