200mL of 0 . 2MHCl is mixed with 300mL of 0 . 1MNaOH . The molar...

$200mL$ of $0 . 2MHCl$ is mixed with $300mL$ of $0 . 1MNaOH .$ The molar heat of neutralization of this reaction is $- 57 . 1kJ .$ The increase in temperature in $C ∘$ of the system on mixing is $x \times {10}^{- 2} .$ The value of $x$ is (Nearest integer)

[Given: Specific heat of water $= 4 . 18J{g}^{- 1}{K}^{- 1}$

Density of water $= 1 . 00g{cm}^{- 3}$ ]

(Assume no volume change on mixing)

Correct answer

Step-by-step solution

milimole $\mathop{HCl}_{\mathop{40}_{0}} + \mathop{NaOH}_{\mathop{30}_{0}} ⟶ \mathop{NaCl}_{\mathop{-}_{30}} + \mathop{{H}_{2}O}_{\mathop{-}_{-}}$ ${\Delta H}_{\text{neut}} = - 57 . 1KJ$

$\Delta H = \left[- 57 . 1 \times 30 \times {10}^{- 3} \times {10}^{3}\right]J = 1713J$

$q = m \cdot s . \Delta T$

$1713 = 500 \times 4 . 18 \times \DeltaT$

$\Delta T = 0 . 8196K = 81 . 96 \times {10}^{- 2}K \approx 82 \times {10}^{- 2}K$

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