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Molarity of aqueous glucose `(C_(6)H_(12)O)^(6)` will be, if mole fraction of glucose is 0.4.A. 10MB. 3.7MC. 0.4MD. 2M |
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Answer» `n_("glucose")=0.4rArrw_("glucose")=0.4xx180=72(M wt. C_(6)H_(12)O_(6)=180)` `n_(H_(2)O)=0.6rArrW_(H_(w)O)=0.6xx18=18.8` `W_("solution")=72+10.8=82.8gm` `d_("solution")=(82.8)/(V_("solution"))=2.07 gm//mlrArrV_("solution")=(82.8)/(2.07)=40ml` Molarity `rArrM=(0.4)/(40)xx1000=10M` |
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