Question

How many grams of BaCl2 are needed to prepare 100cm³ of a 0.250M Bacl2 ? please answer this... ​

Answer 1

Explanation:

atomic mass of bacl2 is 208

Answer 2

Answer:

52×10⁻⁴ grams of BaCl₂ are needed to prepare 100cm³ of a 0.250M BaCl₂.

Explanation:

• The molarity of the solution is defined as the total number of moles of solute per liter of the given solution. Which is calculated as,

                                                  $\mathrm{M=\frac{n}{V}}$          (1)

Where,

M=molarity of the solution

n=number of moles of the solute

V=volume of the solution

And the number of moles of the solute is calculated as,

                                              $\mathrm{n=\frac{m}{M} }$               (2)

Where,

m=given mass of the solute

M=molar mass of the solute

From the question we have,

The volume of the solution(V)=100cm³=100×10⁻⁶L

The molarity of the solution(M)=0.250

The molar mass of BaCl₂=208 gram

When the necessary values are entered into equation (2), we obtain;

$\mathrm{n=\frac{m}{208} }$      (3)

Equation (1) is obtained by inserting equation (3) and the necessary values;

$\mathrm{0.250=\frac{m}{208} \times \frac{1}{100\times 10^{-6}}}$

$\mathrm{m=0.250\times 208\times 100\times 10^{-6}}$

$\mathrm{m=52\times 10^{-4}\ gram}$

Hence,52×10⁻⁴ grams of BaCl₂ are needed to prepare 100cm³ of a 0.250M BaCl₂.

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