What weight of glucose (mol.wt. `= 180`) would have to be added to `1700 g` of water at `20^(@)C` to lower its vapour pressure `0.001 mm`? The vapour pressure of pure water is `17 mm Hg` at `20^(@)C`.

What weight of glucose (mol.wt. `= 180`) would have to be added to `17

1.	What weight of glucose (mol.wt. `= 180`) would have to be added to `1700 g` of water at `20^(@)C` to lower its vapour pressure `0.001 mm`? The vapour pressure of pure water is `17 mm Hg` at `20^(@)C`.
Answer» Correct Answer - 1 Given, `P^(@) - P_(S) = 0.001 mm`, `P^(@) = 17 mm, P_(S) = 16.999, m = 180`, `M = 18, W = 1700, w = ?` Now `(P^(@)-P_(S))/(P_(S)) = (w//m)/(W//M)` `rArr (0.001)/(16.999) = (w xx 18)/(180 xx 1700) rArr w = 1 g`

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What weight of glucose (mol.wt. `= 180`) would have to be added to `1700 g` of water at `20^(@)C` to lower its vapour pressure `0.001 mm`? The vapour pressure of pure water is `17 mm Hg` at `20^(@)C`.

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