Answer:
159 g OF LiNO2 WILL BE USED TO MAKE 0.25 L OF 0.75 M SOLUTION.
Step-by-step explanation:
How many grams of LiNO2 are required to make 250 mL of 0.75 M?
First calculate the molarity in mol per dm3
0.75 M of LiNO2 reacts in 250 mL = 250 /1000 L volume
0.75 M = 0.25 L
In 1 L, the molarity of LiNO2 will be:
= (0.75 * 1/ 0.25) M
= 3 mol/dm3 of LiNO2
Next is to calculate the molarity in g/dm3:
Molarity in mol/dm3 = molarity in g/dm3 / RMM.
RMM of LiNO2
(Li = 7 , N =14 , 0 = 16)
RMM = ( 7 + 14 + 16 * 2) = 53 g/mol
Molarity in mol/dm3 = Molarity in g/dm3 / RMM
Molarity in g/dm3 = Molarity in mol/dm3 * RMM
Molarity in g/dm3 = 3 * 53
Molarity in g/dm3 = 159 g/dm3.
So therefore, to make 250 mL of 0.75 M of LiNO2, we use 159 grams of LiO2.