When 400 mL of 0.2 M H2SO4 solution is mixed with 600 mL of 0.1 M NaOHsolution, the increase in temperature of the final solution is x×10−2KThe value of x (Round off to the Nearest Integer) is [Use :H+(aq)+OH−(aq)→H2O:ΔγH=−57.1 kJ mol−1Specific heat of H2O=4.18J k−1g−1Density of H2O=1.0 g cm−3[ Assume no change in volume of solution on mixing.]

When 400 mL of 0.2 M H2SO4 solution is mixed with 600 mL of 0.1 M NaOH

1.	When 400 mL of 0.2 M H2SO4 solution is mixed with 600 mL of 0.1 M NaOHsolution, the increase in temperature of the final solution is x×10−2KThe value of x (Round off to the Nearest Integer) is [Use :H+(aq)+OH−(aq)→H2O:ΔγH=−57.1 kJ mol−1Specific heat of H2O=4.18J k−1g−1Density of H2O=1.0 g cm−3[ Assume no change in volume of solution on mixing.]
Answer» When $400 m L$ of $0.2 M H_{2} S O_{4}$ solution is mixed with $600 m L$ of $0.1 M N a O H$ solution, the increase in temperature of the final solution is $x \times 10^{- 2} K$ The value of x (Round off to the Nearest Integer) is [Use : $H^{+} (a q) + O H^{-} (a q) \to H_{2} O : Δ_{γ} H = - 57.1 {kJ mol}^{- 1}$ $Specific heat of H_{2} O = 4.18 {J k}^{- 1} g^{- 1}$ $Density of H_{2} O = 1.0 {g cm}^{- 3}$ [ Assume no change in volume of solution on mixing.]