#### Answer

(a) $-2.1 \times 10^2 kJ $
(b) $-33kJ$

#### Work Step by Step

(a)
1. Divide the amount (in mols) by the coefficient of $H_2O(l)$:
$\frac{34.2}{ 1} = 34.2 mol - reaction$
2. Find the necessary heat:
** Since the water freezes, the reaction is reversed, so the $\Delta_rH^{\circ}$ is negative.
$34.2mol-reaction \times -6.0 kJ/mol-reaction = -2.1 \times 10^2 kJ $
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(b)
Molar Mass ($H_2O$):
1.008* 2 + 16* 1 = 18.016g/mol
1. Calculate the amount in moles:
$100g \times \frac{1 mol (H_2O)}{ 18.016} = 5.551 mol$
2. Divide the amount by the coefficient of $H_2O(l)$:
$\frac{5.551}{ 1} = 5.551 mol - reaction$
3. Find the necessary heat:
** Since the water freezes, the reaction is reversed, so the $\Delta_rH^{\circ}$ is negative.
$5.551mol-reaction \times -6.0 kJ/mol-reaction = -33kJ$