Question

How many moles of nacl are required to prepare 0.80 l of 6.4 m nacl? 0.13 mol nacl 5.1 mol nacl 7.2 mol nacl 8.0 mol nacl?

Answer 1

NaCl = n/V 

6.4M = n/0.80L 

n= 5.12mol

Answer 2

Answer:

5.12 moles of NaCl are required to prepare 0.80 L of 6.4 M NaCl.

Explanation:

Molarity (M) is a concentration measure that indicates the number of moles of solute that are dissolved in a given volume.

The Molarity is then determined by:

$Molarity (M)=\frac{number of moles of solute}{volume}$

Molarity is expressed in units$\frac{moles}{liter}$.

6.4 M NaCl indicates that 1 liter of solution there are 6.4 moles of NaCl.

You can apply a rule of three as follows: if in 1 liter of solution there are 6.4 moles of NaCl, in 0.8 L how many moles are there?

$moles=\frac{0.8 L*6.4 moles}{1 L}$

moles=5.12

5.12 moles of NaCl are required to prepare 0.80 L of 6.4 M NaCl.