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Molarity of H_(2)SO_(4) is 0.8 and its density is "1.06 g/cm"^(3). What will be its concentration in terms of molality and mole fraction ? |
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Answer» Solution :Molarity of `H_(2)SO_(4)=0.8` MEANS 0.8 MOLE `H_(2)SO_(4)` are present in 1 L of the solution `"0.8 mole "H_(2)SO_(4)=0.8xx98 g = 78.4g,"1 L "H_(2)SO_(4) " solution "=1000xx1.06g=1060 g` `THEREFORE"Mass solvent (water )"=1060-78.4 g = 981.6g=(981.6)/(18)" moles"="54.53 moles"` Thus, 981.6 g of water contain 0.8 mole of `H_(2)SO_(4)`. `therefore"Molality "=("0.8 mol")/("981.6 g")xx"1000 g kg"^(-1)="0.815 mol kg"^(-1)` Further, the solution contains 0.8 mole of SOLUTE in 54.53 moles of solvent. `therefore"Mole FRACTION of solute"=(0.8)/(0.8+54.53)=0.014` |
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