Question

A 4.939 g sample of sodium nitrate is dissolved in 50.084 g of distilled water, initially at 20.0C.

The temperature of the water fell to 14.5C after the sample dissolved. Find

a. The heat of solution in joules per gram of sodium nitrate.

b. The heat of solution in joules per mole of sodium nitrate.

Answer 1

m = 4.939 NaNO3

m = 50.084 water

dT= 14.5-20 = -5.5°C

a)

heat of solution per gram

total mass = 4.939 +50.084

Assume Cp similar to water since 90% is water

Q = m*Cp*dT = 55.023*(4.184)*(5.5) = -1266.189276 J

then, since this is for a

m = 4.939 g of NaNO3

Hsoln = -Q/m = 1266.189276/4.939 = 256.365 J/g (positive since it is endothermic)

MW of NANO3 = 84.9947

mol = mass/MW = 4.939/84.9947 = 0.05810950565

Hsoln = -Q/n = 1266.189276/0.0581095= 21789.710 J/mol (positive since it is endothermic)