Question

How many grams of lithium nitrate, LiNO3 (68.9 g/mol), are required to prepare 342.6 mL of a 0.783 M LiNO3 solution ? A) 0.00389 g B) 18.5 g C) 0.0541 g D) 30.1 g E) 0.00635 g

Answer 1

Answer:

b) 18.5 g

Explanation:

The first step is the calculation of number of moles of $LiNO_3$ with the molarity equation, so:

$M=\frac{#~mol}{L}$

With the volume in "L" units (342.6 mL= 0.342 L ) we can find the moles, so:

$0.783~M\frac{#~mol}{0.342~L}$

$#~mol=~0.783*0.342=0.268~mol~LiNO_3$

Now, we have to calculate the molar mass to convert the moles to grams. For this we have to know the atomic mass of each atom:

Li ( 6.94 g/mol)

N (14 g/mol

O (16 g/mol)

With these values and the number of atoms in the molecule we can do the math:

(6.94 x 1) + ( 14 x 1) + (16 x 3 ) = 68.94 g/mol

The final step is the calculation of the grams, so:

$0.268~mol~LiNO_3~\frac{69.94~g~LiNO_3}{1~mol~LiNO_3}=~18.5~g~LiNO_3$

We will need 18.5 g of LiNO3 to do a 0.783 M solution with a volume 342.6 mL.