what volume ( in mL) of a 0.680 M KBr solutionwould you need if you want 4.00 grams of KBr?
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.
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**
Math is correct, just thought I saw 6M, not 0.680M and yes the last zero should be taken as a sig fig. Apologies.
To determine the volume of a 0.680 M KBr solution needed to obtain 4.00 grams of KBr, we can use the formula:
moles = mass / molar mass
First, calculate the number of moles of KBr using the given mass:
molar mass of KBr (potassium bromide) = atomic mass of K (potassium) + atomic mass of Br (bromine)
= 39.10 g/mol + 79.90 g/mol
= 119.00 g/mol
moles = mass / molar mass
= 4.00 g / 119.00 g/mol
≈ 0.0336 moles
Now, we can use the formula for molarity to find the volume:
moles = molarity * volume
volume = moles / molarity
volume = 0.0336 moles / 0.680 M
≈ 0.0494 L
Finally, convert the volume from liters to milliliters:
volume = 0.0494 L * 1000 mL/L
= 49.4 mL
Therefore, you would need approximately 49.4 mL of a 0.680 M KBr solution to obtain 4.00 grams of KBr.
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.