When 1 mol of sodium nitrate, is dissolved in 1 cubic decimeter of water, 40 kJ of heat energy is absorbed. What is the drop in temperature when 17.0g of sodium nitrate is dissolved in 1 cubic decimeter of water? [Relative atomic mass:N=14, O=16, Na=23, specific heat capacity of water=4J/g °C and density of solution = 1.0 gcm-3] ]

1.0 ° C

Explanation:

The molar mass for Sodium Nitrate NaNO₃ = (23+14 + (16*3)) = 85

Number of moles of NaNO₃ = mass of NaNO₃ / molar mass of NaNO₃

⇒ 17/85 = 1.38 moles

Since 1 mole of NaNO₃ dissolved in 1 cubic decimeter of water, 40 kJ of heat energy is absorbed.

when 1.38 mole of NaNO₃ dissolved in 1 cubic decimeter of water, x kJ of heat energy is absorbed ...

Then; x kJ of 1.38 mole of NaNo₃ = 1.38 * 40 kJ = 55.2 kJ of heat absorbed.

Using the relation : Q = mcΔT to determine the temperature drop; we get:

55.2 = 17 * 4 (ΔT)

55.2 = 68 ΔT

ΔT = 0.8 ° C

ΔT ≅ 1.0 ° C

Therefore, the drop in temperature when 17.0g of sodium nitrate is dissolved in 1 cubic decimeter of water is 1.0 ° C