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+

Question

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+

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)`.

Monjur Alahi

5 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

Correct Answer - 6
`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`

5 views Answered 2023-02-05 15:29