Question

A student determines the heat of dissolution of solid
**ammonium bromide** using a coffee-cup calorimeter of
negligible heat capacity.

When **6.34** g of
**NH _{4}Br**(s) is dissolved in

ΔH_{dissolution} = kJ/mol

Answer #1

no of moles of NH4Br = W/G.M.Wt

= 6.34/97.94 = 0.0647moles

mass of solution = 6.34+ 119 = 125.34g

q = mcT

= 125.34*4.184*(22.76-25)

= -1174.7J

ΔH_{dissolution } =
-1174.7J

= -1174.7/0.0647 = -18.156KJ/mole >>>>>answer

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