# Answer to Question #57776 in General Chemistry for Tony lundy

Question #57776

A bathroom cleaning is, by weight, 5.8% citric acid (H3C6H5O7, 192.1235g/mol), 8.75% dipropylene glycol butoxy ether (C9H20O3, 190.28288g/mol), 842 PPM limonene (C10H16, 136.2364g/mol) and the rest water.

The density of the butoxy ether is 0.907g/mL and the density of the solution is 1.042g/cm3. Calculator the concentration of citric acid in m and N, the butoxy ether in M and Volume % and the limonene in X and m. Report value a minimum of 4 decimal places.

X (H3C6H5O7)=

Vol% (C9H20O3)=

M (C10H16)=

N (H3C6H5O7)=

m (C9H20O3)=

X (C10H16)=

The density of the butoxy ether is 0.907g/mL and the density of the solution is 1.042g/cm3. Calculator the concentration of citric acid in m and N, the butoxy ether in M and Volume % and the limonene in X and m. Report value a minimum of 4 decimal places.

X (H3C6H5O7)=

Vol% (C9H20O3)=

M (C10H16)=

N (H3C6H5O7)=

m (C9H20O3)=

X (C10H16)=

Expert's answer

Weight of 1 liter of solution is

msol = 1000*1.042 = 1042 g

The weight of the components contained in 1 liter of solution:

m(H3C6H5O7) = 5.8*1042/100 = 60.43600 g

m (C9H20O3) = 8.75*1042/100 = 91.17500 g

m (C6H16) = 842*1042/100000 = 0.87736 g

m (H2O) = 1042 – (60.43600 + 91.17500 + 0.87736) = 889.51164 g

The number of moles of the components in one liter of solution:

n(H3C6H5O7) = 60.43600/192.1235 = 0.31457 mol

n(C9H20O3) = 91.17500/190.28288 = 0.479155 mol

n (C6H16) = 0.87736/136.2364 = 0.006440 mol

n (H2O) = 889.51164/18.01528 = 49.37540 mol

n(Σ) = 0.31457 + 0.479155 + 0.006440 + 49.37540 = 50.17556 mol

Calculate:

X (H3C6H5O7) = n(H3C6H5O7)*100/n(Σ) = 0.31457*100/50.17556 = 0.62694 %

X (C10H16) = n(C10H16)*100/n(Σ) = 0.006440*100/50.17556 = 0.012835%

M (C10H16) = n (C6H16)/Vsol = 0.006440/1 = 0.006440 M

Citric acid is tribasic, so

N (H3C6H5O7) = 3n(H3C6H5O7)/ Vsol = 3*0.31457/1 = 0.94371N

V(C9H20O3) = m(C9H20O3)/ρ(C9H20O3) = 91.17500/0.907 = 100.52370 ml

Vol% (C9H20O3) = V(C9H20O3)*/ Vsol = 100.52370*100/1000 = 10.05237%

Answer:

X (H3C6H5O7) = 0.62694 %

Vol% (C9H20O3) = 10.05237%

M (C10H16) = 0.006440 M

N (H3C6H5O7) = 0.94371N

m (C9H20O3) = 91.17500 g

X (C10H16) = 0.012835%

msol = 1000*1.042 = 1042 g

The weight of the components contained in 1 liter of solution:

m(H3C6H5O7) = 5.8*1042/100 = 60.43600 g

m (C9H20O3) = 8.75*1042/100 = 91.17500 g

m (C6H16) = 842*1042/100000 = 0.87736 g

m (H2O) = 1042 – (60.43600 + 91.17500 + 0.87736) = 889.51164 g

The number of moles of the components in one liter of solution:

n(H3C6H5O7) = 60.43600/192.1235 = 0.31457 mol

n(C9H20O3) = 91.17500/190.28288 = 0.479155 mol

n (C6H16) = 0.87736/136.2364 = 0.006440 mol

n (H2O) = 889.51164/18.01528 = 49.37540 mol

n(Σ) = 0.31457 + 0.479155 + 0.006440 + 49.37540 = 50.17556 mol

Calculate:

X (H3C6H5O7) = n(H3C6H5O7)*100/n(Σ) = 0.31457*100/50.17556 = 0.62694 %

X (C10H16) = n(C10H16)*100/n(Σ) = 0.006440*100/50.17556 = 0.012835%

M (C10H16) = n (C6H16)/Vsol = 0.006440/1 = 0.006440 M

Citric acid is tribasic, so

N (H3C6H5O7) = 3n(H3C6H5O7)/ Vsol = 3*0.31457/1 = 0.94371N

V(C9H20O3) = m(C9H20O3)/ρ(C9H20O3) = 91.17500/0.907 = 100.52370 ml

Vol% (C9H20O3) = V(C9H20O3)*/ Vsol = 100.52370*100/1000 = 10.05237%

Answer:

X (H3C6H5O7) = 0.62694 %

Vol% (C9H20O3) = 10.05237%

M (C10H16) = 0.006440 M

N (H3C6H5O7) = 0.94371N

m (C9H20O3) = 91.17500 g

X (C10H16) = 0.012835%

## Comments

## Leave a comment