Question

when a 3.00 g sample of KCN is dissolved in water in a calorimeter that has a total heat capacity of 1.419 kJ . K^(-1), the temperature decreases by 0.380 K. Calculate the molar heat of solution of KCN

Answer 1

so a 3 g sample of KCN made the calorimeter increase in temperature (0.38 K), how much heat does that equal?

\(heat=q=m*C_p*(0.38~K)\)

Answer 2

after you found the heat produced by 3 g of that sample, you need to find the heat produced by a mole of that sample. Set up a ratio:

\(\dfrac{q_{sample}}{3~g}=\dfrac{q_{mole}}{Molar~mass~of~KCN}\)

Answer 3

btw, i forgot to include a negative sign in the first equation, as exothermic processes are negative in sign.

Answer 4

so 1.61766 would be the heat of the sample... correct?

Answer 5

the i would divide that number by the molar mass which is 65.12 which gives me an answer of 0.0248. would that be my final answer?

Answer 6

i made a mistake, there was no need to use the mass because the mass of the calorimeter is already taken into account.

so it was only: \(q=C_p*\Delta T\)

Answer 7

the second step doesn't sound right, can you type what you did?

Answer 8

1) q= 1.419x.380

so q= 0.53922
2) 0.53922/ 65.12

65.12 is the molar mass of KCN and i got a "final" answer of 0.00828. is that correct? sorry this problem is complicated

Answer 9

the first part is good, but you're not taking into account the whole ratio:

\(\dfrac{0.53922~J}{3~g}=\dfrac{q_{mole}}{65.12~g/mol}\rightarrow q_{mole}=\dfrac{(0.53922~J)*(65.12~g/mol)}{3~g}\)

Answer 10

i just noticed another mistake i made, the units of q should be in kJ, not J.

so it should read "0.53922 \(\color{red}{kJ}\)"