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A 4 % solution of sucrose C_(12)H_(22)O_(11) is isotonic with 3 per cent solution of an unknown organic substance. Calculate the molar mass of the unknown substance. |
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Answer» Solution :Since the two solutions are isotonic, they must have same concentrations in moles/litre. For sucrose solution, we gave`""` Concentration `=4g//100cm^(3)"(GIVEN) = 40 g/litre "=(40)/(342)" moles/litre"` `""(because" Molar mass of sucrose, "C_(12)H_(22)O_(11)="342 g mol"^(-1))` For unknown substance, suppose M is the mol. mass. Then Concentration = `3g//100cm^(3)"(Given)= 30 g/litre"=(30)/(M) "moles/litre"` THUS, we have `(30)/(M)=(40)/(342) or M=(30xx342)/(40)="256.5 g mol"^(-1)` |
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