Question

how much bacl2 would be needed to make 250 ml of a solution having same concentration of cl- as the one containing 3.78g of nacl per 100 ml?

Answer 1

Let's take the case of NaCl.

NaCl->Na+ + Cl-

[Cl-]=[NaCl]=3.78*1000/58.5*100=?

{Molarity=weight(in g)*1000/Mol.mass(in g/mol)*volume}

Now, BaCl2.

BaCl2->Ba2+ + 2Cl-

[Cl-]=?

[BaCl2]=[Cl-]/2=?

[Cl-]/2=w*1000/208.3*250

{Mol. mass of BaCl2=208.3, Ba=137.3, Cl=35.5}

w=[BaCl2]*208.3*250/1000=?

Answer 2

Answer: 16.81 grams of $BaCl_2$ is dissolved in 250 mL of water.

Explanation:

Molarity of NaCl = $\frac{\text{mass of NaCl}}{\text{molar mass of NaCl}\times \text{volume is liters}}=\frac{3.78 g}{58.5 g/mol\times 0.1 L}=0.6461 M$

The concentration chloride ions in 0.1 L NaCl solution is 0.6461 mol/l .

$BaCl_2\rightarrow Ba^{2+}+2Cl^-$

1 mole of $BaCl_2$ produced on mole of barium ion and 2 mole of chloride ions.

For 0.6461 mol/L concentration of chloride ion then $[BaCl_2]$

$[BaCl_2]=\frac{0.6461 mol/L}{2}=0.3230 mol/L$

Molarity of $BaCl_2$:

$0.3230 mol/L=\frac{\text{mass of }BaCl_2}{\text{molecular mass of }BaCl_2\times \text{volume is liters}}$

$=\frac{x g}{208.23 g/mol\times 0.25 L}$

x = 16.81 grams

16.81 grams of $BaCl_2$ is dissolved in 250 mL of water.