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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. 31.54B. 6.3 uC. 630 uD. 63 u |
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Answer» Correct Answer - D `pi_(x)=pi_("glucose")` `iC_(x)RT=iC_("glucose")RT` `C_(x)=C_("glucose")` `(1.05)/(MM)xx(1000)/(100)=(3g)/(180)xx(1000)/(100)` `MM=63` |
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