Answer to Question #170415 in Chemistry for Hema

Question

How many milliliters (ml) of 1 M MgSO4 must be added to 50 ml of a 0.1 M MgSO4 solution to produce 1000 ml with a final MgSO4 concentration = 0.75 M? How much water?

Answer 1

Total number of moles of the final solution equals:

n(total) = V(final) × M(final) = 1,000 mL × 0.75 M = 1L × 0.75 M = 0.75 mol

The initial solution contains:

n(initial) = V(initial) × M(initial) = 50 mL × 0.1 M = 0.05 L × 0.1 M = 0.005 mol

From here, the number of moles of the solute that must be added is:

n(added) = n(final) - n(initial) = 0.75 mol - 0.005 mol = 0.745 mol

As a result, the volume of 1M solution to be added equals:

V(added) = n(added) / M(added) = 0.745 mol / 1M = 0.745 L = 745 mL

Finally, the volume of water that must be added equals:

V(water) = 1 L - 0.005 L - 0.745 L = 0.25 L = 250 mL

Answer: 745 mL of 1M solution and 250 mL of water