Question

Instant hot packs work by crystallizing sodium acetate (NaCH3COO). The molar enthalpy of crystallization for sodium acetate is -56.7 kJ/mol. How many grams of sodium acetate are needed to warm 125.0 ml of water from 21 degrees C to 35.4 degrees C?

Answer 1

$V = 125.0ml$

Since, $m = \rho × V = 1g/ml × 125ml = 125g$

$∆T = T_2 - T_1 = 35.4°C - 21°C = 14.4°C$

$c = 4.186J/g°C$

H = mc∆T

H = 125 × 4.186 × 14.4 = 7534.8J = 7.535kJ

Since the molar enthalpy of crystallization for sodium acetate is -56.7 kJ/mol

56.7kJ = 1 mole of sodium acetate

7.535kJ = x moles

x = $\dfrac{7.535×1}{56.7}$ = 0.133 moles

1 mole of $CH_3COONa$ = 82g

0.133 moles = xg

x = 0.133 × 82 = 10.9g

This means that 10.9 grams of sodium acetate is needed to warm 125.0 ml of water from 21°C to 35.4°C.