Question

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] ]​

Answer 1

1.0 ° C

Step-by-step 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