Question

how many grams of Barium Chloride are needed to prepare 100 cm or 0.2 50 M Barium Chloride solution

Answer 1

Molecular mass of BaCl2 = 208 g

Let the mass of BaCl2 required = w

Moles of BaCl2 = w/208

Molarity=   [nos of moles of solute *100] /volume of solution

0.250 = [w *1000] / [208 *100 ]

w = [0.250 *208 *100 ] / 1000

=5.2 g ans

Answer 2

Thus 5.2 g of Barium Chloride is required to prepare 100 cm molar solution.

Explanation:

We are given that:

Molarity of solution = 100 cm or 0.250 M

Solution:

Molecular mass of BaCl2 = 208 g

Let the mass of BaCl2 required = x

Moles of BaCl2 = x / 208

Molarity =   [Number of moles of solute x 100] / Volume of solution

0.250 = [x x 1000] / [208 x 100 ]

x = [0.250 x 208 x 100 ] / 1000

x = 5200 / 1000

x =5.2 g

Thus 5.2 g of Barium Chloride are required to prepare 100 cm molar solution.