Question

Moles, molarity, and composition

a) How many gmol are there in 21.2 kg of isobutyl succinate (C12H22O4)?
b) If sucrose (C12H22O11) crystals flow into a hopper at a rate of 4.5 kg s^-1, how many gram-moles of sucrose are transferred in 30 min?
c) A solution of 75 mM tartaric acid (C4H6O6) in water is used as a standard in HPLC analysis. How many
grams of C4H6O6 ·H2O are needed to make up 10 ml of solution?
d) An aqueous solution contains 60 μM salicylaldehyde (C7H6O2) and 330 ppm dichloroacetic acid
(C2H2Cl2O2). How many grams of each component are present in 250 ml? You can assume the density of
the solution to be same as water, as 1 g/mL (this is an appropriate assumption, since this is a very dilute
solution).

Answer 1

a)

Molecular mass of isobutyl succinate , (C12H22O4 ) = 230

No. of gm. moles = Given mass of C12H22 O4 in gms. / Molecular mass of C12H22 O4

............................= (21.2 x 1000) /230

.......................... = 92.17 gm.moles

b)

Given , .. sucrose which flows into hopper in 1 sec. = 4.5 Kgs

the weight of sucrose that flows into hopper in 1 sec. = ( 4.5 x 1000 )gms

the weight of sucrose that flows through the hopper in 30 min.   = ( 4.5 x1000 x 30 x 60 ) gms

so the number of gram moles of sucrose transfered in 30 min. ........= ( 4.5 x 1000 x30 x 60 ) / .................................................................................................................mol. mass of sucrose

...........................................................................................................= (4.5 x 1000 x 30 x 60 ) ........................................................................................................................................../342

...........................................................................................................= 23684 .21 gm . moles

....................................................................................................or, = 2.3684 x 10⁴ gm. moles

c)

75mM of tartaric acid (C4H6O6) = ( 75 /1000 ) = 0.075 M of Tartaric acid

Since ,the weight of C4 H6 O6 .H2O required to prepare 1000 ml of 1M C4H6 O6 solution = ..........................................................................................Molecular mass of .C4H6 O6 .H2O

................................................................................................................................= 144.0gms.

-----------------------------------------------------------------------------------------0.075M--------------------= .....................................................................................................................( 144 x 0.075 )gms.

--------------------------------------------------------------10.0 ml of 0.075 M-------------------------------= ..........................................................................................................(144 x 0.075 x10.0 )/1000 ...............................................................................................................................= 0.108 gms

......................................................................................................................= 1.08 x 10^-1 gms.

d)

60 micro moles of C7H6O2 = 60 x 10^-6 M

............................................. =( 60 x 10^-6 x 122 ) gm. / L

............................................. = 7.32 x 10^-3 gms / L

so the weight of C7H6 O2 in 250 ml = 1.83 x 10^-3 gms.

330 ppm of C2H2 Cl2 O2 = 330 mgm. per litre

......................................... = 0.330 gms / L

......................................... = 0.330 / 129

........................................... = 2.558 x 10^-3  gm. / L

so weight of C2H2Cl2 O2 in 250 ml = 6.395 x 10^-4 gms.

Moles of water ................... = 250/ 18

............................................ = 13.88 gm .moles