What mass of glucose (in grams) is added to 72 g of water which lower vapour pressure by 2%?
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What mass of glucose (in grams) is added to 72 g of water which lower vapour pressure by 2%?
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Given:
Mass of water = 72 gm
The relative lowering of vapour pressure = 2%
To Find:
The mass of glucose required for this lowering.
Calculation:
-Let no of moles of glucose be n1
- No of moles of water, n2 = 72/18 = 4
- We know that:
Relative lowering of vapour pressure = Mole fraction of solute
⇒ 2/100 = n1/(n1+n2)
⇒ 0.02 = n1/(n1+4)
⇒ n1 = 0.02 n1 + 0.08
⇒ 0.98 n1 = 0.08
⇒ m/180 = 0.08/0.98
⇒ m = 180 × 0.08/0.98
⇒ m = 14.4/0.98
⇒ m = 14.7 gm
- So, 14.7 gm of glucose is added to 72 g of water which lower vapour pressure by 2%.
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