In: Chemistry
1.a) Suppose a solution of 14.8M NH3. How many milliliters of this solution are required to prepare 100 mL
of 1.0M NH3, when diluted?
b) A stock solution of Nitric acid is 15.8 M. How many milliliters of the stock solution is required to make
up to 1.00 L of 0.12`M Nitric acid?
1.a)
Use the formula C1V1 =C2V2
Where, C1 = initial concentration = 14.8M
V1 = initial volume = ?
C2 = final concentration = 1.0M
V2 = final volume = 100 ml
C1V1 =C2V2 Thus
V1 = C2V2 /C1
Substitute the value in above equation
V1 = 1.0100/14.8 = 6.7568 ml
6.7568 milliliterl of 14.8 M NH3 stock solution required
b)
Use the formula C1V1 =C2V2
Where, C1 = initial concentration = 15.8M
V1 = initial volume = ?
C2 = final concentration = 0.12M
V2 = final volume = 1.00 liter = 1000 ml
C1V1 =C2V2 Thus
V1 = C2V2 /C1
Substitute the value in above equation
V1 = 0.121000/15.8 = 63.29 ml
63.29 milliliterl of 15.8 M Nitric acid stock solution required