Answer
(a) $$\Delta H = -58 \ kJ/mol$$
(b)
We could improve the precision of this measurement by adding a greater quantity of potassium hydroxide. With greater values, we get a greater temperature and more significant figures. But, it should not be much, since the assumption that the specific heat of solution is equal to the same of water depends on a low concentration of solute.
Work Step by Step
(a)
Mass of solution = 55.9 g + 0.205 g = 56.105 g
q = m * specific heat * $\Delta T$
1. Calculate $q_{water}$:
q = ( 56.105 )( 4.18 )( 24.4 - 23.5 )
q = $ 211 $ J = 0.211 kJ
$q_{KOH} = -q_{water} = -0.211$ kJ
2. Find the $\Delta H$
$ KOH $ : ( 1.008 $\times$ 1 )+ ( 16.00 $\times$ 1 )+ ( 39.10 $\times$ 1 )= 56.11 g/mol
$$ 0.205 \space g \times \frac{1 \space mole}{ 56.11 \space g} = 3.65 \times 10^{-3} \space mole$$
$$\Delta H = \frac{-0.211 \ kJ}{3.65 \times 10^{-3} \ mole} = -58 \ kJ/mol$$
(b)
We could improve the precision of this measurement by adding a greater quantity of potassium hydroxide. With greater values, we get a greater temperature and more significant figures. But, it should not be much, since the assumption that the specific heat of solution is equal to the same of water depends on a low concentration of solute.
