1.

0.0125 mol of sucrose is dissolved in 100 gm of water and it undergo partial inversion according to following equation C_12H_22O_11+H_2OtoC_6H_12O_6+C_6H_12O_6 If elevation in boiling point of solution is 0.104^@C calculate 1/10 mol percentage of sugar inverted (K_b,_(H_2O)=0.52).

Answer»


SOLUTION :`underset("Mol")(C_12H_22O_11)+underset(0.0125)(H_2O)tounderset(0)(C_6H_12O_6)+underset(0)(C_6H_12O_6)`
`0.0125-x " " x ""x`
`DeltaT_b=m_1K_b+m_2K_b+m_3K_b`
`m_1+m_2+m_3=0.104/0.52=0.2`
`(0.0125-x+x+x)/100 xx1000=0.2`
x=0.0075
mol % `=0.0075/0.0125xx100=60%`
`1/10th` of mol% =`60/10=6`


Discussion

No Comment Found

Related InterviewSolutions