Question

How many mL of a 0.045 M BaCl2 contain 15g of BaCl2 (208 g mol–1
). Detailed answer please

Answer 1

Answer:

Givens:

0.45 M solution of BaCl2

To determine the volume of the BaCl2 solution that would contain

15.0 grams, you set up the proportion:

Solution:

1. 0.45 x 208 = 93.6 grams of BaCl2 per liter.

2.1000 mL/93.6 gm = x mL/15.0 gm. -> X = 160.3 mL

3.Do this : 160.3 mL x 0.45 gm/mL =

0.072 moles x 208 gm/mol =

15.00 gm.

Explanation:

I hope this helped.

Answer 2

Answer:

160.3 mL of a 0.045 M BaCl2 contain 15g of BaCl2 (208 g mol–1 )

Explanation:

Given,0.45 M solution of BaC$l_2$

We want to find volume of the BaC$l_2$solution that would contain15.0 grams.

By creating proportion,we can solve this problem.

We are setting up the proportion,

$\frac{2.1000 \: ml}{93.6 \: gm} = \frac{x \: ml}{15.0 \: gm}$

So,X=160.3 mL

Therefore, 169.3 mL of a 0.045 M BaC$l_2$

contain 15g of BaC$l_2$(208 g mol–1)

Extra information:Barium chloride ( Bac$l_2$) is an inorganic compound . Bac$l_2$ is one of the most common water-soluble salts of barium. Like most other water-soluble barium salts, it is white, highly toxic, and gives a yellow-green color to the flame. It is also hygroscopic, first converting to the dihydrate BaC$l_2$(H2O)₂.

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