# Question 903e4

Jun 7, 2017

The enthalpy of solution is B. -57.61 kJ/mol.

#### Explanation:

There are two heats involved in this problem.

$\text{heat of solution of KOH + heat absorbed by water} = 0$
$\textcolor{w h i t e}{m m m m m} {q}_{1} \textcolor{w h i t e}{m m m m m l l} + \textcolor{w h i t e}{m m m m m} {q}_{2} \textcolor{w h i t e}{m m m m m l} = 0$
color(white)(mmmm)nΔHcolor(white)(mmmmmmmml) + mCΔT color(white)(mmmml)= 0

Let's calculate the heats separately.

n = 13.90 color(red)(cancel(color(black)("g KOH"))) × "1 mol KOH"/(56.11 color(red)(cancel(color(black)("g KOH")))) = "0.2477 mol KOH"

q_1 = nΔH = 0.2477ΔH color(white)(l)"mol"

$m = \text{(13.90 + 122.40) g" = "136.30 g}$
$C = \text{4.184 J·g·°C"^"-1}$
ΔT = "(45.73 - 20.70) °C" = "25.03 °C"

q_2 = mCΔT = 136.30 color(red)(cancel(color(black)("g"))) × "4.184 J"·color(red)(cancel(color(black)("g"^"-1"·"°C"^"-1"))) × 25.03 color(red)(cancel(color(black)("°C"))) = "14 274 J" = "14.274 kJ"

q_1 + q_2 = 0.2477ΔH color(white)(l)"mol" + "14.274 kJ" = 0

ΔH = "-14.274 kJ"/"0.2477 mol" = "-57.62 kJ/mol"#