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Ostrovityanka [42]
2 years ago
15

What is the weight of 4.30 mole of sodium

Chemistry
1 answer:
Dovator [93]2 years ago
3 0

98.857g Na

To find this, set up an equation as so:

4.3 mol Na = \frac{22.99g Na}{1 mol Na}

Multiply everything on the top and divide that by everything on the bottom to get the final answer of 98.857g Na.

Hope this helps!

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First, find the volume the solution needs to be diluted to in order to have the desired molarity:
You have to use the equation M₁V₁=M₂V₂ when ever dealing with dilutions.

M₁=the starting concentration of the solution (in this case 2.6M)
V₁=the starting volume of the solution (in this case 0.035L)
M₂=the concentration we want to dilute to (in this case 1.2M)
V₂=the volume of solution needed for the dilution (not given)

Explaining the reasoning behind the above equation:
MV=moles of solute (in this case KCl) because molarity is the moles of solute per Liter of solution so by multiplying the molarity by the volume you are left with the moles of solute.  The moles of solute is a constant since by adding solvent (in this case water) the amount of solute does not change.  That means that M₁V₁=moles of solute=M₂V₂ and that relationship will always be true in any dilution.

Solving for the above equation:
V₂=M₁V₁/M₂
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That means that the solution needs to be diluted to 75.8mL to have a final concentration of 1.2M.

 Second, Finding the amount of water needed to be added:
Since we know that the volume of the solution was originally 35mL and needed to be diluted to 75.8mL to reach the desired molarity, to find the amount of solvent needed to be added all you do is V₂-V₁ since the difference in the starting volume and final volume is equal to the volume of solvent added.
75.8mL-35mL=40.8mL
40.8mL of water needs to be added

I hope this helps.  Let me know if anything is unclear.
Good luck on your quiz!
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