How many grams of LiNO2 are required to make 250 mL of 0.75 Molarity solution?

159 g OF LiNO2 WILL BE USED TO MAKE 0.25 L OF 0.75 M SOLUTION.

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.

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