Answer:
28.20 mL of the stock solution.
Explanation:
Data obtained from the question include the following:
Molarity of stock solution (M1) = 12.1 M
Volume of diluted solution (V2) = 350.0 mL
Molarity of diluted solution (M2) = 0.975 M
Volume of stock solution needed (V1) =..?
The volume of stock solution needed can be obtained by using the dilution formula as shown below:
M1V1 = M2V2
12.1 x V1 = 0.975 x 350
Divide both side by 12.1
V1 = (0.975 x 350)/12.1
V1 = 28.20 mL.
Therefore, 28.20 mL of the stock solution will be needed to prepare 350.0 mL of 0.975 M HCl solution.
Using the law of dilution :
Mi x Vi = Mf x Vf
2.00 x Vi = 0.15 x 100.0
2.00 x Vi = 15
Vi = 15 / 2.00
Vi = 7.5 mL
hope this helps!
Answer:
The molarity is 2M
Explanation:
First , we calculate the weight of 1 mol of NaCl:
Weight 1mol NaCl= Weight Na + Weight Cl= 23 g+ 35, 5 g= 58, 5 g/mol
58,5 g---1 mol NaCl
233,772 g--------x= (233,772 g x1 mol NaCl)/58,5 g= 4 mol NaCl
<em>A solution molar--> moles of solute in 1 L of solution:</em>
2 L-----4 mol NaCl
1L----x0( 1L x4mol NaCl)/4L =2moles NaCl---> 2 M
Answer:
oxygen and silicon, both are common
