At standard pressure, $C H_{4}$ $CH_4$ boils at 112 K and $H_{2} O$ $H_2O$ boils at 373 K. What accounts for the higher boiling point of $H_{2} O$ $H_2O$ at standard pressure?

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At standard pressure, $C H_{4}$ $CH_4$ boils at 112 K and $H_{2} O$ $H_2O$ boils at 373 K. What accounts for the higher boiling point of $H_{2} O$ $H_2O$ at standard pressure?

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Answer 1 · 2016-12-31T09:10:38

Intermolecular force differentiates..........

Explanation:

Between water molecules, hydrogen bonding operates, which is a potent intermolecular force. The $H - O$ $H-O$ bond is quite polar, i.e. $^{δ +} H - O^{δ -} - H^{δ +}$ $""^(delta+)H-O^(delta-)-H^(delta+)$ , and the dipoles can align. No such intermolecular force operates between methane molecules, (because the $C - H$ $C-H$ bond is reasonably non-polar), and the boiling point of methane is substantially reduced with respect to water.

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At standard pressure, $C H_{4}$ $CH_4$ boils at 112 K and $H_{2} O$ $H_2O$ boils at 373 K. What accounts for the higher boiling point of $H_{2} O$ $H_2O$ at standard pressure?

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At standard pressure, CH4CH_4 boils at 112 K and H2OH_2O boils at 373 K. What accounts for the higher boiling point of H2OH_2O at standard pressure?

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At standard pressure, $C H_{4}$ $CH_4$ boils at 112 K and $H_{2} O$ $H_2O$ boils at 373 K. What accounts for the higher boiling point of $H_{2} O$ $H_2O$ at standard pressure?