If 3.0 liters of oxygen gas react with excess carbon monoxide at STP, how many liters of carbon dioxide can be produced under th
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Answer : The suspension base added will be, 250 mL
Explanation :
As we are given that the initial concentration of solution is, 10 mg/5mL or we can say that 2 mg/mL.
Volume of solution = 250 mL
First we have to determine the mass of PAXIL.
As, 1 mL of solution contain mass of PAXIL = 2 mg
So, 250 mL of solution contain mass of PAXIL =
Now we have to determine the final solution required.
As we are given the final concentration of solution is, 0.1 % w/v that means 0.1 g in 100 mL of solution.
Concentration of solution =
Let the 'x' mL of the suspension base is added. So, the ratio of mass and volume will be:
Thus, the suspension base added will be, 250 mL