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`        calculate the molality molarity mole fraction of KI if the density 20% mass by mass aqeous KI is 1.202g/mol`
2 years ago

```							Dear student Molar mass of KI = 39 + 127 = 166 g mol-120% (mass/mass) aqueous solution of KI means 20 g of KI is present in 100 g of solution.That is,20 g of KI is present in (100 - 20) g of water = 80 g of waterTherefore, molality of the solution = moles of KI/mass of water in Kg  = (20/166)/ 0.08= 1.506 m= 1.51 m (approximately)(b) It is given that the density of the solution = 1.202 g mL?1mass/ density =Volume of 100 g solution 100g/ 1.202 gml-1 = 83.19 mL= 83.19 × 10?3 LTherefore, molarity of the solution = (20/166) / 83.19 × 10?3 L= 1.45 M(c) Moles of KIMoles of water = 80/18 = 4.44Therefore, mole fraction of KI = moles of KI/ moles of KI + moles of water = 0.12/0.12 + 4.44= 0.0263 = 0.0263
```
2 years ago
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