Question #544e6

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Question #544e6

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Answer 1 · 2016-05-02T18:12:51

0.954g

Explanation:

Here 200mL0.5M HCl contaims
$\frac{200 \cdot 0.5}{1000} = 0.1$ $(200*0.5)/1000=0.1$ mole HCl
=0.1gmequivalent HCl

25mLpartly neutralised HCl requires 20.5mL 0.5M NaOH .
forcomlere neutralisation

200mLpartly neutralised HCl requires $20.5 \cdot \frac{200}{25} = 164 m L$ $20.5*200/25=164mL$ 0.5M NaOH for complete neutralisation

164mL0.5M NaOH solution contains
$164 \cdot \frac{0.5}{1000} = .082$ $164*0.5/1000=.082$ mole or gm-equivalent NaOH.

Let the mass of $N a_{2} C O_{3}$ $Na_2CO_3$ added be x g or $\frac{x}{53}$ $x/53$ gm-equivalent
where equivalent mass of $N a_{2} C O_{3} = 53$ $Na_2CO_3=53$
Here total no.of gm-equivalent of base =total no.of gm-equivalent
of acid

$\frac{x}{53} + 0.082 = 0.1$ $x/53+0.082=0.1$

$x = 0.018 \cdot 53 = 0.954 g$ $x=0.018*53=0.954g$

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Question #544e6

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