Question

A)What volume of 0.491 M BaCl2 solution is required to obtain 0.694 mol of BaCl2?

B)What volume of 0.376 M BaCl2 solution is required to obtain 0.725 mol of Cl−(aq)?

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Answer 1

A)

Number of moles = Volume of solution (in L) * Molarity (M)

Moles Required BaCl2 = 0.694 mol

Molarity of solution = 0.491M

0.694 = 0.491 * Volume of solution (in L)

Volume of solution (in L) = 1.41344L = 1413.44 mL

B)

Number of moles = Volume of solution (in L) * Molarity (M)

BaCl2 ------> Ba(2+) + 2Cl-

Moles Required BaCl2 = Half of moles of Cl- = 0.725/2 = 0.3625 moles

Molarity of solution = 0.491M

0.3625 = 0.376 * Volume of solution (in L)

Volume of solution (in L) = 0.96409L = 964.09 mL