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If 1 .71 g of sugar (molar mass = 342) a...

If 1 .71 g of sugar (molar mass = 342) are dissolved in `500 cm^(3)` of solution at 300 K, what will be its osmotic pressure ?

Text Solution

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The correct Answer is:
`0.249`
`W_(B) = 1 . 71 g, M_(B) = 342 g mol^(-1)`
`V = 500 cm^(3) = 0 . 5 L`
`T = 300 K `
`R = 0 . 0 83 L b a r K^(-1) mol^(-1) , pi = ?`
`pi = (W_(B))/( M_(B)) (RT)/(V)`
`= (1 . 71 g xx ( 0 . 0 83 L b a r K^(-1) mol^(-1)) xx 300 K)/( 342 g mol^(-1) xx 0 . 5 L)`
`= (171 xx 3 xx 0 . 083)/( 342 xx 0 . 5 )` bar = 0 . 249 bar
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