KNO3 (s) - -->ž K + (aq) + NO3 - (aq) This reaction was carried out in a Styrofoam insulated calorimeter and the following data were recorded:Mass of solid KNO3 dissolved10.1 gMass of aqueous solution (c = 4.18 J/gºC) 100. gT initial30.0ºCT final21.6ºCMolar mass of KNO3101 g/molIf the mass of KNO3 solid dissolved were doubled while all other experimental conditions were kept the same, what change would occur in delta T, J per reaction, J/g of KNO3, and kJ/mol KNO3? Larger delta T, larger J/reaction, larger J/g, no change kJ/molLarger delta T, no change J/reaction, no change J/g, no change kJ/molLarger delta T, larger J/reaction, no change J/g, no change kJ/molLarger delta T, larger J/reaction, larger J/g, larger kJ/mol

Larger ΔT, larger J, no change J/g, no change kJ/mol is the correct answer.

Explanation:

When one sees data in this question one tends to think that to be solved we need to perform calculations, and that is not the case here.

What we need is to remembember what are properties which depend on quantities (extensive properties) and the concepts and formulas for heat.

Δ Hrxn = - Q cal

Q cal = m x c x ΔT where

m = mass of water in the calorimeter

c = specific heat of water, and

ΔT = change in temperature

ΔT is directly proportional to Q

ΔH rxn is an extensive quantity dependent on the amount of KNO₃, as is J / reaction.

J/g, kJ/mol are intensive properties the moment they are defined as the heat per gram, and heat per mol released or absorbed.

mass of KNO₃ ⇒ doubles heat of reaction, doubles ΔT

Therefore,

c) Larger ΔT, larger J, no change J/g, no change kJ/mol is the correct answer.