Assume you recovered 0.120 ml of citral in this experiment. how many grams of lemon grass oil did you begin with if citral comprises 5.72% of the oil?

1.88 g

Explanation:

Considering:-

Given that:-

% Citral = 5.72 %

Volume of the citral = 0.120 mL

So,

Volume of the grass oil sample = 2.10 mL

Also, Density of the lemon grass oil = 0.893 g/mL

Mass = Density*Volume = 0.893 g/mL*2.10 mL = 1.88 g

1.88 g of lemon grass oil did you begin with if citral comprises 5.72% of the oil.

1.84 g/mL

Explanation:

We know that

percentage of citral =

⇒

solving for V_sample we get 2.09 mL

Now density of the lemon grass oil = 0.88g/mL at room temp.

then mass in gm= Volume×Density= 2.09×0.88 =1.84 gram.

a saline water is fed to a reverse osmosis unit at flow rate o rotameter connected to reject (retentate) line reads 8 installed in reject and permeate line reads 15.2 milli siemens/cm and 300 micro siemens/com respectively.