Question

How much \[\ce{BaCl2 * 2H2O}\] and pure water are to be mixed to prepare 50 g of 12.0% (by mass) BaCl₂ solution?

Options

• 40.4 g

• 42.9 g

• 52.7 g

• 50.0 g

Answer 1

42.9 g

Explanation:

w = 12 g of BaCl₂, W = 100 g of solution

For 50 g of solution, w = 6 g of BaCl₂

W = 50 g of solution

\[\therefore \ce{w_{BaCl_2  \phantom{.}· \phantom{.} 2 H_2O}}\] = `((6 xx 244)/208)` g

= 7.038 g

`"w"_("H"_2"O") = (50 - 7.038)` g

= 42.96 g ≈ 42.9 g