$(a)\;20\%,22\%,70\%\qquad(b)\;22\%,23\%,69.4\%\qquad(c)\;21\%,21\%,69.57\%\qquad(d)\;24\%,22\%,68.5\%$

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On heating $NaHCO_3$ decomposes to gives $CO_2$

$2NaHCO_3\rightarrow Na_2CO_3+H_2O+CO_2\uparrow$

Meq of $NaHCO_3$=Meq of $CO_2$

$\large\frac{w_{NaHCO_3}\times 1000}{84}=\frac{56}{22400}$$\times 2\times 1000$

$w_{NaHCO_3}=0.42g$ in 2.0g mixture

$\Rightarrow \large\frac{0.42}{2}$$\times 100$

$\Rightarrow 21\%$

Now suppose in 1.6g mixture,'x'g of Nacl,0.336g(21% of 1.6)$NaHCO_3$ and rest $Na_2CO_3$ then

Wt of $Na_2CO_3=1.6-0.336-x$

$\Rightarrow (1.264-x)g$

Since both $Na_2CO_3% and $NaHCO_3$ react with HCl and therefore

Meq of $Na_2CO_3$+Meq of $NaHCO_3$=Meq of HCl

$\large\frac{1.264-x}{53}$$\times 1000+\large\frac{0.336}{84}$$\times 1000=25\times 1$

$x=0.151$

Weight of NaCl=0.151g and % of NaCl=9.43%

Weight of $NaHCO_3$=0.336g and % of $NaHCO_3$=21.00%

Weight of $Na_2CO_3=1.264-0.151$

$\Rightarrow 1.113$ and % of $Na_2CO_3=69.57\%$

Hence (c) is the correct answer.

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