If your lab partner (it's always the lab partner) had not heated the sample long enough, leaving some KCIO3 unreacted, your results for standard molar volume would be: a) not affected
b) too low
c) too high

not affected

Explanation:

Oxygen gas is obtained by heating KClO3 in the presence of a catalyst. The amount if KClO3 that decomposed to yield oxygen is obtained by measuring the volume of oxygen gas evolved or by stoichiometry calculation. The equation of the reaction is shown as; 2 KClO3(s). 2 KCl(s) + 3 O2(g).

The Standard Molar Volume is the volume occupied by one mole of any gas at STP. All gases occupy a volume of 22.4 L at STP. The standard molar volume of a gas is constant and is not affected by the extent of reaction of the KClO3. Hence the standard molar volume is not affected.

in the given question, the partial pressure is calculated by using the ideal gas equation, that is, pv = nrt

here, p is pressure, v is volume = 25 l, n is number of moles = 0.0104 moles, r is gas constant = -0.0821 l.atm / mol.k

t is temperature = 273 k

p = nrt / v

p = (0.0104 mol × 0.0821 l.atm / mol.k × 273 k) / 25 l

= 0.0093 atm