Question

A 4.86 kg sample of a solution of the solute chloroform in the solvent (C2H5)2O that has a concentration of 1.91 pph by mass chloroform is available. Calculate the amount (g) of CHCl3 that is present in the sample.
Molar Mass (g/mol)
CHCl3 119.38
(C2H5)2O 74.12

Density (g/mL):
CHCl3 1.483
(C2H5)2O 0.7138

Name/Formula:
chloroform
CHCl3
diethyl ether
(C2H5)2O

Answer 1

pph = parts per hundred = %.
1.91 pph = 1.91%
4.86 kg = 4860 g sample; of which there is 1.91% CHCl3. Therefore, the amount of CHCl3 is
4,860 g x 0.0191 = ??
Check my thinking. Check my work.

Answer 2

5. The tetraethyl lead [Pb(C2H5)4] in a 25.00-mL sample of aviation gasoline was shaken with 15.00mL of 0.02095M I2