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The mole fraction of a solute in a solution is 0.1. At 298 K, molarity of this solution is same as its molality Density of the solution at 298 K is "2.0 g cm"^(-3). The ratio of the molecular weights of the solute and solvent ((MW_("solute"))/(MW_("solvent"))), is |
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Answer» Solution :Mole fraction of solute = 0.1 means moles of solute = 0.1, moles of solvent = 0.9 `"Mass of solute "=0.1xxM_("solute")G,` `"Mass of solvent "=0.9xxM_("solvent")g` `=(0.1M_("solute")+0.9M_("solvent"))g` Total VOLUME of solution `=("Mass")/("Density")=(0.1M_("solute")+0.9M_("solvent"))/(2xx1000)L` Molarity of solution `=(0.1xx2000)/(0.1M_("solute")+0.9M_("solvent"))"mol L"^(-1)` `=(200)/(0.1M_("solute")+0.9M_("solvent"))` `"Molality of solution"` `=(0.1xx1000)/(0.9M_("solvent"))"mol KG"^(-1)=(100)/(0.9M_("solvent"))` `"As molarity = molality"` `(200)/(0.1M_("solute")+0.9M_("solvent"))=(100)/(0.9M_("solvent"))` `"or"0.1M_("solute")+0.9M_("solvent")=2xx0.9M_("solvent")` `"or"(0.1M_("solute"))/(M_("solvent"))+0.9=1.8` `"or"(M_("solute"))/(M_("solvent"))=(0.9)/(0.1)=9` |
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