200 mL of 0 . 2 M HCl is mixed with 300 mL of 0 . 1 M NaOH . The molar heat of neutralization of this reaction is - 57 . 1 kJ . The increase in temperature in C ∘ of the system on mixing is x × 10 - 2

Question

200 mL of 0 . 2 M HCl is mixed with 300 mL of 0 . 1 M NaOH . The molar heat of neutralization of this reaction is - 57 . 1 kJ . The increase in temperature in C ∘ of the system on mixing is x × 10 - 2 . The value of x is (Nearest integer) [Given: Specific heat of water = 4 . 18 J g - 1 K - 1 Density of water = 1 . 00 g cm - 3 ] (Assume no volume change on mixing)

TestHub · Accepted Answer

milimole HCl 40 0 + NaOH 30 0 ⟶ NaCl - 30 + H 2 O - - ΔH neut = - 57 . 1 KJ ΔH = - 57 . 1 × 30 × 10 - 3 × 10 3 J = 1713 J q = m · s . ΔT 1713 = 500 × 4 . 18 × Δ T ΔT = 0 . 8196 K = 81 . 96 × 10 - 2 K ≈ 82 × 10 - 2 K

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