Question

when a 2.00 g sample of rbbr is dissolved in water in a calorimeter that has a total heat capacity of 0.541 kj·k⁻¹, the temperature decreases by 0.490 k. calculate the molar heat of solution of rbbr.
δh_soln = kj/mol

Explanation:

Step1: Calculate the heat absorbed by the calorimeter

The heat absorbed by the calorimeter \(q\) is given by the formula \(q = C\times\Delta T\), where \(C\) is the heat capacity of the calorimeter and \(\Delta T\) is the change in temperature.
Given \(C = 0.541\space kJ\cdot K^{-1}\) and \(\Delta T=0.490\space K\)
\(q = 0.541\times0.490\space kJ\)
\(q=0.26509\space kJ\)

Step2: Calculate the number of moles of \(RbBr\)

The molar mass of \(RbBr\) is \(M = 85.47 + 79.904=165.374\space g/mol\)
The mass of \(RbBr\) is \(m = 2.00\space g\)
The number of moles \(n=\frac{m}{M}=\frac{2.00\space g}{165.374\space g/mol}\approx0.0121\space mol\)

Step3: Calculate the molar heat of solution \(\Delta H_{soln}\)

The molar heat of solution \(\Delta H_{soln}=\frac{q}{n}\)
Substitute \(q = 0.26509\space kJ\) and \(n = 0.0121\space mol\)
\(\Delta H_{soln}=\frac{0.26509\space kJ}{0.0121\space mol}\approx21.9\space kJ/mol\)

Answer:

\(21.9\space kJ/mol\)