Answer:
(a, b)
(a) \[{{M}_{NaOH}}=\frac{{{w}_{B}}\times
1000}{{{m}_{B}}\times V}=\frac{20\times 1000}{40\times
1000}=2.5\,mol\,{{L}^{-1}}\]
(b) \[{{M}_{KCl}}=\frac{n}{V}\times
1000=\frac{0.5}{200}\times 1000=2.5mol\,{{L}^{-1}}\]
(c) \[{{M}_{NaOH}}\frac{{{w}_{B}}\times
1000}{{{m}_{B}}\times V}=\frac{40\times 1000}{40\times
100}=10\,mol\,{{L}^{-1}}\]
(d) \[{{M}_{KOH}}=\frac{{{w}_{B}}\times
1000}{{{m}_{B}}\times V}=\frac{20\times 1000}{56\times
200}-1.785mol\,{{L}^{-1}}\]
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