The concentration of the stock solution she needs is 100 milli molar (mm) and she needs to make 1.2 milli liters (ml) solution of the drug-a. the drug is available in a salt form with a molecular weight of 181.6 grams / mole. what is the amount (quantity in grams) of drug-a will she have to weigh-out in order to make the stock solution

Molarity is defined as the number of moles of solute in 1 L of solution 
molarity of stock solution to be prepared - 100 x 10⁻³ mol/L
volume of stock solution to be prepared - 1.2 mL 
Therefore number of moles in 1.2 mL - 100 x 10⁻³ mol/L x 1.2 x 10⁻³ L 
number of moles of drug - 1.2 x 10⁻⁴ mol 
mass of drug required - 1.2 x 10⁻⁴ mol x 181.6 g/mol = 21. 8 mg
21.8 g of drug is required to make the stock solution

Answer : The volume of in Reaction 1 = 240.45 ml

The volume of in Reaction 2 = 480.89 ml

Solution : Given,

Temperature =         ()

Pressure = 0.900 atm

The mass of = 1 gram

Molar mass of = 84.007 g/mole

The given reactions are,

Reaction 1 :

Reaction 2 :

Calculation for Reaction 1 :

First we have to calculate the moles of .

Moles of =

From the Reaction 1, we conclude that

2 moles of gives 1 mole of

0.0119 moles of gives of

Using ideal gas equation :  

where,

P = pressure of gas

V = volume of gas

n = Number of moles

T = temperature of gas

R = gas constant = 0.0821 L atm/mole K

Now put all the given values in ideal gas law, we get the volume of

The volume of in Reaction 1 is 240.45 ml

Calculation for Reaction 2 :

The moles of = 0.0119 moles

From the Reaction 2, we conclude that

1 moles of gives 1 mole of

So, the moles of = the moles of = 0.0119 moles

Using ideal gas equation,

Now put all the given values in ideal gas law, we get the volume of

The volume of in Reaction 2 is 480.89 ml