A group of students carried out a Boyle\'s Law experiment as done in your lab wo

Question

A group of students carried out a Boyle's Law experiment as done in your lab work by using a computer. In the Table below are the results of volume and pressure measurements. Is this relationship p and V Inverse or Direct In the above experiment volume was changed from 5.8mL (P = 1.91 atm as above) to 15.8mL Calculate the new pressure. Write gas law equation and rearranged Eq and calculation. Table values Gas law and Calculation Steps (With units) Rear. Eq Fill in the blank by using the correct word In law as the pressure (P) of a sample of a gas increases the temperature (T) of the gas if the volume is kept constant. A sample of 1.350 g of an impure sample of KHP was titrated with 0.120 M NaOH standardized solution. The volume of NaOH needed to neutralize was 35.50 ml. of NaOH. Calculate the present of KHP in this sample. (Molar mass of KHP is 204.2g). Show all steps as 1.2, 3 etc. with units and c.v. and f.v as done in the lab.

Explanation / Answer

The volume remained constant in both the experiments ( from 5- 5.8 ml = 0.8ml and 19 to 19.8 ml =0.8m). But the pressrue has changed. So Pressure is independent of volume.

2.From Boyles law, PV= constant, hence 5.8*1.91= P2*15.8 l

P2= 5.8*1,91/15.8 =0.7011ml

3. from PV= nRT. At constant volume P varies as absoulte temperature. So with an increase in pressure at constant volume, temperature also increases.

4. The reaction between KHP and NaOH is KHC8H4O4(aq) + NaOH(aq) --> KNaC8H4O4(aq) + H2O(l)

moles of NaOH in 35ml of 0.12M= 0.12*35/1000=0.0042

1 mole of NaOH requires 1 mole of NaOH as per the stoichiometric equation

0.0042 moles of NaOH requires 0.0042 moles of KHP

moles of KHP =mass of KHP/ Molar mass

mass of KHP= moles of KHP* molar mas = 0.0042*204.2 =0.857 gm

mass of original sample = 1.35 gm

Mass % of sample = 100*0.857/1.35=63.48%

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