Gram (+) and Gram (-) forms of bacteria are differentible through staining with

Options

A. 44g
B. 20.33g
C. 22g
D. 58.66g

Answer

D

Explanation

Molar mass of `CH_(3)COOH=` =60gr/mole
Molar mass of `C_(2)H_(5)OH=` 46 gr/mole
Molar mass of `CH_(3)COOC_(2)H_(5)`= 88gr/mole
`[CH_(3)COOH]_("Initial")`
`=(60)/(60 xx 5)=0.2" mole/L"`
`[CH_(3)COOC_(2)H_(5)]_("eqm")`
`=(44)/(88 xx 5)=0.1" mole/L"`
`CH_(3)COOH+C_(2)H_(5)OH Leftrightarrow `
`{:(,"initial",0,2m,0,2m),(,"Ateq",(0.2-0.1),0.1m)`
`CH_(3)COOC_(2)H_(5)+H_(2)O`
`0.1m " "0.1m`
`K=([CH_(3)COOC_(2)H_(5)][H_(2)O])/([CH_(3)COOH] [C_(2)H_(5)OH])`
`K=(0.1 xx 0.1)/(0.1 xx 0.1) =1`
In second case
`[CH_(3)COOH]_("initial")=0.4m`
`[CH_(2)H_(5)OH]_("initial")=0.2m`
If x is the amount of acid and alcohol reacted
`[CH_(3)COOH]_(eqm)=(0.4-x)m`
`[C_(2)H_(5)OH]_(eqm)=(0.2-x)m`
`[CH_(3)COOC_(2)H_(5))_(eqm)=[H_(2)O)]_(eqm)=xm`
`K=(x^(2))/((0.4-x)(0.20-x))=1, x=(8)/(60)m`
moles of ethyl aetate produced `=8/60 xx 5`
mas of ethyl acetate Produced `=2/3 xx 88 =58.66`