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Mole fraction of solute in aqueous solution of 30% NaOH

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2 years ago

aparazita sharma
41 Points
```							let mass of solution be 100 g then mass of naoh is 30 g and mass of water is 100-30= 70 gnow moles of naoh = mass of naoh / molar mass of naoh  =  30 / 40 = 0.75similarly moles of water = 70 /18 =3.88so total moles=0.75+3.88=4.63mole fraction of solute = moles of naoh / total moles                                        0.75/ 4.63=0.16 or 27/ 167
```
2 years ago
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