Question

calculate the volume strength of 1.6 n h2o2 sol​

Answer 1

Answer

We know that strength = Normality × Eq. ...Strength of 1.6 N H2O2 solution = 16 × 17 g/1.Now 68 g of H2O2 gives 22400 ml O2 at NTP.16 × 17 g of H2O2 will give 2240068 × 1.6 × 17 = 8960 ml of O2 at NTP.But 1.6 × 17 g of H2O2 are present in 1000 ml of H2O2 solution.

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Answer 2

Correct question:

Calculate the volume strength of the 1.6 n $H_{2} O_{2}$ solution.

The volume strength of the 1.6 n $H_{2} O_{2}$ is 8.96 volume.

Step-by-step explanation:

To find:

Find the volume strength of the 1.6 n $H_{2} O_{2}$.

Formula used: strength = Normality × Eq. wt.

Solution:

Eq. wt of $H_{2} O_{2}=17$

So the volume strength of the 1.6 n $H_{2} O_{2}$ $=1.6\times17gL^{-1}$

Now 68 g of $H_{2} O_{2}$ gives 22400 $mLO_{2}$ at NTP/STP.

$16 \times 17 g$ of $H_{2} O_{2}$ will give$\frac{22400}{68}\times1.6\times17 = 8960$  ml of $O_{2}$ at NTP/

But$1.6 \times 17 g$ of $H_{2} O_{2}$ are present in $1000 ml$ of $H_{2} O_{2}$ solution.

Hence $1000 ml$ of $H_{2} O_{2}$ solution gives $8960 ml$ of $O_{2}$ at STP.

$1ml$ of $H_{2} O_{2}$ solution will give $=\frac{8960}{1000}$

$=8.96ml$ of $O_{2}$ at NTP.

Hence, the volume strength of the 1.6 n $H_{2} O_{2}$ solution is 8.96 volume.

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