Question

Trial Caleulation 1 need the following data: R. 8.314 kPa L mol"K! T from the information above in Kelvin V from the information above in L lP .P in kPa Note: the units all have to be in the same form when used in calculations!!! H:O: density is 1.02 g/mL RT 5 mL of hydrogen peroxide was decomposed at 25 °C releasing 48 mL of gas. The temperature of the container used to measure the volume of gas produced was 20 °c. To find BOTH the concentration of hydrogen peroxide in the solution AND the percentage of hydrogen peroxide in solution work through the following steps. 1. Rearrange the ideal gas law to solve for n. 2T 2. Convert the temperature of the water in the collection vessel from C to Kelvin (K). Therefore: T 25 oc+273-248 K Convert the current barometric pressure in the room from hPa to kPa. The Macquarie university weather station data can be used to find the pressure http:/laws.mq,.edu.au. Use the current pressure which is given in hectopascals. 3. 72? kya / / ? 10 hPa 1 kPa So divide hPa by 10 to convert to kPa. 4nd O.OL O Sna 4. Convert the volume of oxygen from mL to Liters, : o.coSL hot You are now ready to solve for the number of moles of O2. Be sure the units cancel so that you end up with only the moles of O2 Answer the questions in the table on the following page and show your calculations. Note: Each question requires the answer to the previous question and some additional information to answer the question. Page 13

barometric pressure 101 kpa. Is my working out correct? can you please solve this. Thankyou.

Answer 1

Well done! You have solved the first three problems accurately. I have checked them thoroughly.

(iv) The no. of moles of H2O2 = 4.2*10^-3 mol

The final volume = 5+5 = 10 mL = 10*10^-3 L

Therefore, the concentration of H2O2 used in the reaction = 4.2*10^-3 mol/10*10^-3 L

= 0.42 mol/L or 0.42 M

v) The molar mass of H2O2 = 34 g/mol

Therefore, the mass of H2O2 in the original 5 mL sample = 4.2*10^-3 mol * 34 g/mol = 0.143 g

vi) The density of H2O2 = 1.45 g/mL

Therefore, the volume of H2O2 in the original 5 mL sample = 0.143 g/(1.45 g/mL) ~ 0.1 mL

vii) The % volume of H2O2 in the total 5 mL sample = (0.1 mL/5 mL) * 100 = 2%