\( 5.5 g \) of sodium hydroxide is dissolved in \( 175 mL \) of water. Using a coffee-cup calorimeter,

Question

8. \( 5.5 g \) of sodium hydroxide is dissolved in \( 175 mL \) of water. Using a coffee-cup calorimeter, the temperature change of the water is measured to be \( +2.10^{\circ} C \). The specific heat capacity of water is \( 4.184 J / g ^{\circ} C \). What is the thermochemical equation for this process? (A) \( NaOH (s) \rightarrow NaOH ( aq )+1.54 kJ \) (B) \( NaOH ( s )+1.54 kJ \rightarrow NaOH ( aq ) \) (C) \( NaOH (s) \rightarrow NaOH ( aq )+11.2 kJ \) (D) \( NaOH ( s )+11.2 kJ \rightarrow NaOH ( aq ) \)

Answer 1

We have given-----

Mass of NaOH = 5.5 g

Volume of water = 175 ml

Temperature change of water = +2.10°C

specific Heat capacity of water = 4.184 J/g°C

\(\because\) Temperature of water increases, it means, dissolution of NaOH in water is a exothormic process.

We know that,

density of water = 1 g/ml at STP

\(\therefore\) Mass of water = 175 ml x 1g/mL

 = 175 g

Heat absorbed by water Q = m x s x Δt

where, m = mass of water

S = specific Heat of water  = 4.18 J/g°C

 Δt = change in temperature

\(\therefore\) Q = 175 g x 4.18 4J/g°C x 2.10°C

= 1.537 KJ

 = 1.54 KJ

\(\therefore\) The correct Thermochemical equation will be----

NaOH(s) → NaOH(aq) + 1.54 KJ

Hence, Option (A) is right answer.