In the given reaction: NaOH (.aq.)+HCl (.aq.) arrow NaCl (.aq.)+H2O (.l.) 50 ml of HCl containing 7.3 g of HCl per litre and 100 ml solution of NaOH containing 4 g NaOH per litre reacts, at any instant 0.5 g of NaCl is formed. Hence, the amount of NaOH which did not react is:

Question

In the given reaction: NaOH (.aq.)+HCl (.aq.) arrow NaCl (.aq.)+H2O (.l.) 50 ml of HCl containing 7.3 g of HCl per litre and 100 ml solution of NaOH containing 4 g NaOH per litre reacts, at any instant 0.5 g of NaCl is formed. Hence, the amount of NaOH which did not react is: (A) 0.06 g (B) 3.66 g (C) 10.8 g (D) 0.63 g

Tardigrade Admin · Accepted Answer

Milli moles of NaOH in sol =10 Milli moles of HCl in sol =10 At any instant millimoles of NaCl formed =(0.5/58.5)× 103≈8.5 ∴ Amount of NaOH left =0.4 -(.0.0085× 40.) = 0.06 g

Q. In the given reaction: NaOH(aq)+HCl(aq)→NaCl(aq)+H2​O(l) 50 ml of HCl containing 7.3 g of HCl per litre and 100 ml solution of NaOH containing 4 g NaOH per litre reacts, at any instant 0.5 g of NaCl is formed. Hence, the amount of NaOH which did not react is:

0.06 g

3.66 g

10.8 g

0.63 g

Milli moles of NaOH in sol =10 Milli moles of HCl in sol =10 At any instant millimoles of NaCl formed =58.50.5​×103≈8.5 ∴ Amount of NaOH left =0.4−(0.0085×40) =0.06g

Milli moles of $N a O H$ in sol $= 10$
Milli moles of $H Cl$ in sol $= 10$
At any instant millimoles of $N a Cl$ formed $= \frac{0.5}{58.5} \times 1 0^{3} \approx 8.5$
$∴$ Amount of $N a O H$ left $= 0.4 - (0.0085 \times 40)$
$= 0.06 g$