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 ºC of the system on mixing is x × 10–2. The value

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 ºC 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–1K–1 Density of water = 1.00 g cm–3] (Assume no volume change on mixing) 0.2 M HCl ds 200 mL dks 0.1 M NaOH ds 300 mL ds lkFk fefJr fd;k x;k gS bl vfHkfØ;k ds fy, mnklhuhdj.k eksyj Å"ek gS –57.1 kJ fefJr djus ij fudk; dk rki x × 10–2 c<+ tkrk gSA x dk eku gS ________. ¼fudrVe iw.kk±d esa½ [dn;k gS% ty dh fof'k"V Å"ek = 4.18 J g–1K–1 ty dk ?kuRo = 1.00 g cm–3] ¼eku yhft,% fefJr djus ij vk;ru esa dksbZ ifjorZu ugha gksrk gS½

Prashant Kotian · Accepted Answer

Correct answer: .. HCl + NaOH  NaCl + H2O Hneut = –57.1 KJ milimole 30 — — 0 30 — H = [–57.1 × 30 × 10–3 × 103]J = 1713 J q = m.s.T 1713 = 500 × 4.18 × T T = 0.8196 K = 81.96 × 10–2 K  82 × 10–2 K

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