what volume ( in mL) of a 0.680 M KBr solutionwould you need if you want 4.00 grams of KBr?

Answer 1

Moles = Molarity x Volume in Liters => Volume (L) = moles KBR / Molarity of Solution
Molarity (given) = 6-Molar in KBr = 6-moles KBr/L soln
Moles KBr in 4.0-g = 4.0-g/119-g/mole = 0.034-mole KBr
Therefore, Vol(L) of 6M solution containing 4-g KBr = 0.034-mole KBr / 6-mol/L soln =0.0056-L x 1000-ml/L = 5.6-ml.

Answer 2

Solve for moles needed to obtain 4.00g:

4.00g*(mole/119.002g)=0.033612 moles

Molarity (M)= moles/Volume (L)

You know moles needed to obtain 4.00g of KBr; you know Molarity of the solution.

Solve for volume:

0.680 M=0.033612 moles/Volume (L)

Volume (L)=0.033612 moles/0.680 M

Volume (L)=0.0494 L or 494 mL

They answer above isn't correct nor has the correct number of significant figures.

Answer 3

I aplogize, there is a typo in the above post.

Look for ** for correction.

Solve for moles needed to obtain 4.00g:

4.00g*(mole/119.002g)=0.033612 moles

Molarity (M)= moles/Volume (L)

You know moles needed to obtain 4.00g of KBr; you know Molarity of the solution.

Solve for volume:

0.680 M=0.033612 moles/Volume (L)

Volume (L)=0.033612 moles/0.680 M

Volume (L)=0.0494 L or 49.4 mL**

Answer 4

Math is correct, just thought I saw 6M, not 0.680M and yes the last zero should be taken as a sig fig. Apologies.