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A solution of substance containing `1.05` g per 100 mL was found to be isotonic with `3%` glucose solution . The molecular mass of the substance is :A. 31.5B. 6.3C. 630D. 63 |
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Answer» Correct Answer - D `(1.05)/(Mw)xx10=(3)/(180)xx10` `Mw=1.05xx60=63` |
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