One liter of $\frac{N}{2} H C l$ solution is heated in a beaker. It was observed that when the volume of the solution was reduced to $600 m l$ , $3.25 g$ of $H C l$ in lost. What is the normality of the new solution?

1.70

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0.685

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0.342

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0.171

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Solution

The correct option is B $0.685$
$N_{i} V_{i} = N_{1} V_{1} + (n \times M) V_{2}$
Where $N_{i}, V_{i}$ are initial normality & volume of $H C l$ solution $N_{1}, V_{1}$ are the normality & volume of solution when it is reduced to $600 m$
$n \times M \to$ Normality of $3.25 g H C l$ lost $(n = 1)$
$V_{2} \to 400 m l$
$M = \frac{3.25 / 36.5}{0.4}$
$∴ \frac{N}{2} \times 1000 = N_{1} \times 600 + \frac{3.25 / 36.5}{0.4} \times 400$
$N_{1} = 0.685 N$

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One liter of N2HCl solution is heated in a beaker. It was observed that when the volume of the solution was reduced to 600ml, 3.25g of HCl in lost. What is the normality of the new solution?

The correct option is B 0.685NiVi=N1V1+(n×M)V2Where Ni,Vi are initial normality & volume of HCl solution N1,V1 are the normality & volume of solution when it is reduced to 600mn×M→Normality of 3.25g HCl lost (n=1)V2→400mlM=3.25/36.50.4∴N2×1000=N1×600+3.25/36.50.4×400N1=0.685N

One liter of $\frac{N}{2} H C l$ solution is heated in a beaker. It was observed that when the volume of the solution was reduced to $600 m l$ , $3.25 g$ of $H C l$ in lost. What is the normality of the new solution?