Question

What mass of ammonium nitrate is required to lower the temperature of 400 mL of
water from 25 oC to 1oC? The Hsoln for ammonium nitrate is 34 kJ/mol.

Answer 1

400 ml = 400 g of water
ΔT = 25 – 1 = 24 o
C
C = 4.18 J/g∙C
Q = cmΔT = 4.18 J/g∙C x 400 g x 24 C = 40128 J = 40.128 kJ
40.128 kJ / 34 kJ/mol = 1.18 mol of ammonium nitrate required
80.043 g/mol x 1.18 mol = 94.45074 g.
Answer: 94.45074 g.