Given is the following information: $H_{2} {SO}_{4} (l) + 5 H_{2} O (l) ⟶ H_{2} {SO}_{4} \cdot 5 H_{2} O (l) : Δ_{1} H^{\circ} = - 58.0 kJ {mol}^{- 1}$ A solution of specific heat capacity $4.184 J K^{- 1} g^{- 1}$ is obtained when 1 mol of H₂ SO₄is added to 5 mol of H₂O. The temperature change in the above process will be about

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

154.0 °C

98.0 °C

73.76 °C

54.2 °C

answer is C.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

We will have $m c_{p} Δ T = - Δ_{f} H^{\circ}$

Where $m = m_{H_{2} {SO}_{4}} + m_{H_{2} O} = (n M)_{H_{2} {SO}_{4}} + (n M)_{water} = (1.0 {mol}^{2} (98 g {mol}^{- 1}) + (5.0 mol) (18 g {mol}^{- 1}) = 188 g$ Where $Δ T = \frac{- Δ_{f} H}{m c_{p}} = \frac{58 \times 10^{3} J}{(188 g) (4.184 J^{\circ} C^{- 1} g^{- 1})} = {73.74}^{\circ} C$