(c) (q = mcΔT =) 25.0 × 4.18 × (20.2 − 12.2) OR 25.0 × 4.18 × 8
(= 836 (J) or 0.836 (kJ))
Not if m = 29
Ignore sign of q
1
4.00 g NH4NO3 = 4.00/80 OR 0.0500 mol
1
ΔH⦵soln = 836/0.05 = 16720 = (+)16.7(2) kJ mol−1
Allow ecf from M1 and/or from M2
–16.7(2) = 2/3
+19.4 = 2/3 (using m = 29 in M1)
−19.4 = 1/3
+2.68 = 2/3
−2.68 = 1/3
+587 or +588 = 2/3
−587 or −588 = 1/3
Allow 2 sig figs or more
1
(d) (2 × 0.1/8) × 100 = 2.5%
Allow ecf from ΔT in (c)
1
(e) use a larger mass/amount of NH4NO3 / solid
Marking points are independent
Allow smaller volume of water / less water
Allow use more NH4NO3
Not larger volume of water
Ignore higher concentration (of NH4NO3)
Ignore any references to changing apparatus e.g. insulation
1
so temperature change/decrease is greater
OR final temperature is lower
Allow temperature increase is greater
Not final temperature is higherr
1
(f) heat gain (from the surroundings) / incomplete dissolvingg
Allow incomplete reaction
Allow thermal energy gain
Not heat loss
Ignore energy gain
Ignore references to mistakes in method
1