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Ainat [17]
3 years ago
11

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
Chemistry
1 answer:
Anika [276]3 years ago
4 0

Answer:

b) 18.5 g

Explanation:

The first step is the <u>calculation of number of moles</u> of LiNO_3 with the molarity equation, so:

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

With the <u>volume in "L" units</u> (342.6 mL= 0.342 L ) we can <u>find the moles</u>, 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 <u>molar mass</u> to convert the moles to grams. For this we have to know the <u>atomic mass</u> 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 <u>calculation of the grams</u>, so:

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

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

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