A solution contains 4 g of Na2CO3 and NaCl- 25 mL of this solution requires 50 mL of N10 HCl for complete neutralization- The - composition of mixture is as nearest integer -

Let mass of nbsp; Na_2CO_3 nbsp;and NaCl in given mixture be 'a' and 'b' g respectively. ∴ nbsp;a+b =4 nbsp; nbsp; nbs ...

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Standard XII

Chemistry

Stoichiometric Calculations

A solution co...

Question

A solution contains $4$ g of ${N a}_{2} C O_{3}$ and $N a C l$ . $25$ mL of this solution requires $50$ mL of $\frac{N}{10}$ $H C l$ for complete neutralization. The $%$ composition of mixture is (as nearest integer) :

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Solution

Let mass of ${N a}_{2} C O_{3}$ and $N a C l$ in given mixture be $^{'} a^{'}$ and $^{'} b^{'}$ g respectively.
$∴ a + b = 4 . . . (i)$
The solution is allowed to react with $H C l$ ; only ${N a}_{2} C O_{3}$ reacts and thus, Meq. of ${N a}_{2} C O_{3} =$ Meq. of $H C l$ in $25$ mL $= 50 \times (\frac{1}{10}) = 5$
$∴$ Meq. of ${N a}_{2} C O_{3}$ in $250$ mL $= \frac{(5 \times 250)}{25} = 50$
or, $\frac{a}{\frac{106}{2}} \times 1000 = 50$
$∴ a = 2.65$
By (i), $b$ $= 4 - 2.65 = 1.35$
$∴ {N a}_{2} C O_{3} = (\frac{2.65}{4}) \times 100 = 66.25$ %
and $N a C l$ $=$ $33.75$ %
So, answer is $34$ %.

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A solution contains 4 g of Na2CO3 and NaCl. 25 mL of this solution requires 50 mL of N10 HCl for complete neutralization. The % composition of mixture is (as nearest integer) :

A solution contains $4$ g of ${N a}_{2} C O_{3}$ and $N a C l$ . $25$ mL of this solution requires $50$ mL of $\frac{N}{10}$ $H C l$ for complete neutralization. The $%$ composition of mixture is (as nearest integer) :