At `27^(@)C,a1.2%` solution (`wt.//"vol.")` of glucose is isotonic with `4.0g//"litre"` of urea solution. Find the molar mass of urea, if the molar ma

Question

At `27^(@)C,a1.2%` solution (`wt.//"vol.")` of glucose is isotonic with `4.0g//"litre"` of urea solution. Find the molar mass of urea, if the molar ma

At `27^(@)C,a1.2%` solution (`wt.//"vol.")` of glucose is isotonic with `4.0g//"litre"` of urea solution. Find the molar mass of urea, if the molar mass of glucose is `180`. Also report the osmotic pressure of solution if `100mL` of each are mixed at `27^(@)C`. (`R=0.082 L` atm `"mol"^(-1)K^(-1)`, Molar mass of glucose `=180g//"mole"`)

Monjur Alahi

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2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Correct Answer - `60 g//"mole", 1.64` atm
`1.2%` solution (`wt.//"vol"`) of glucose is isotonic with `4g//"lite"` of urea solution at `27^(@)C`, so
`C_(1)=C_(2)`
`(1.2xx1000)/(180xx100)=(4)/(M_("urea")`
so `M_("urea")=60 g//"mole"`.
When equal volume (`i.e. 100ml` each) of these two solutions are mixed, the concentration gets halved.
So total concentration `=(1)/(2)((1.2)/(180)+(4xx100)/(60))xx(1000)/(100)=(1)/(15)M`
`therefore` osmotic pressure `P=CRT`
`=(1)/(15)xx0.082xx300=1.64` atm