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`        35 ml sample of hydrogen peroxide gives off 500ml of o2 at 27°C and 1atm pressure. volume strength of h2o2 sample wii be`
10 months ago

Arun
14889 Points
```							Dear Avinash Use PV=nRT and solve for n = PV/RT to get the number of moles of O2 given off and substitute n=PV/RT Substitute P= 1 atm V=.500 L (notice you need to convert to L to match the units in R) R=0.082 L atm/ (mol K) T= 300 K a. Solve for n to get moles of O2 released. n=.0203 b. You can get the volume of O2 relesed at stp from 22.4 L/moles of O2 at STP = 0.4553 c. For every mole of H2O2 you get one mole of O2 released, so that tells you moles H2O2 in the solution = 00203 = 15.48 g H2O2 in the solution If you assume the density of H2O2 is the same as water, then you have 15.5 mL of H2O2 in the solution and the V/V % is 15.5x100/35 = 44% You could also say the w/v % is the same number. It is unclear which you are looking for.  RegardsArun (askIITians forum expert)
```
10 months ago
Dylan
13 Points
```							See, first find volume if oxygen that canbe produced at stp...Therfore, v1/t1=v2/t2(modi 's law)V2=500*273/300(by substituting values in the equation)=455ml35ml can produce 455ml of oxygen at stp,Therfore(unitary method) 1000ml can produce 1000*455/35=13000ml=13l
```
2 months ago
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