1.

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

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`


Discussion

No Comment Found

Related InterviewSolutions