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Calculate osmotic pressure of a solution...

Calculate osmotic pressure of a solution obtained by mixing `100 mL` of `3.4%` solution "(weight/volume)" of urea "(molecular weight 60)" and `100 mL` of `1.6%` solution "(weight/volume)" of cane sugar "(molecular weight 342)" at `20^(@)C`.

Text Solution

Verified by Experts

`pi_(1)V_(1)=W_(1)/(Mw_1)RT_(1)` for urea
`pi_(2)V_(2)=W_(2)/(Mw_2)RT_(2)` for sugar
Since `100 mL` of urea solution is mixed with `100 mL` of cane sugar solution, and thus the total volume becomes `200 mL` in which `3.4 g` urea and `1.6 g` sugar is present.
`:.(pi_(1))xx200/1000=3.4/60 xx0.821 xx 293 =6.82` atm
`:.(pi_(2))xx200/1000=1.6/342 xx0.821 xx 293 =0.56` atm
`:.pi_("total")=pi_(1) + pi_(2)=6.82 + 0.56 = 7.38` atm
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Knowledge Check

  • The osmotic pressure of a solution containing 100 ml of 3.4% solution (w/v) of urea (mol mass 60 g/mole) and 50 ml of 1.6% solution (w/v) of cane-sugar (mol mass 342 g/mole) at 27ºC is:

    A
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    B
    8.98 atm
    C
    17.06 atm
    D
    9.70 atm

  • Calculate Osmotic Pressure of a solution obtained on mixing 100 mL of 3.4% (wt/vol) "solution of area" (Mol.wt 60) "and" 50 mL "of" 1.6% (wt//vol) "solution of canesugar" (Mol.Wt 342) "at" 27^(@)

    A
    9.704
    B
    10
    C
    12.74
    D
    15

  • A 0.6 % solution of urea (molecular weight = 60) would be isotonic with

    A
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    B
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    C
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    D
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