Answer:
The answer is "Option a".
Explanation:
Multiply the mass of the solutions by 3.17% to get the mass of [tex]BaCl_2[/tex]:
[tex]\to 427 \ g \times 0.0317 = 13.56\ g[/tex]
Divide this mass with BaCl2's molecular concentration to acquire the mole ratio:
[tex]molar mass = 208.24 \frac{g}{mol}\\\\moles = \frac{13.56}{208.24} = 0.065 \ moles \\\\[/tex]
[tex]= 6.5 \times 10^{-2} \ moles[/tex]
A 427 g sample is 3.17% by mass barium chloride contains 6.5 × 10⁻² moles of BaCl₂.
A 427 g sample is 3.17% by mass barium chloride, that is, there are 3.17 g of barium chloride per 100 g of sample.
The mass of barium chloride in 427 g of the sample is:
[tex]427gSample \times \frac{3.17gBaCl_2}{100gSample} = 13.5gBaCl_2[/tex]
We can calculate the number of moles in 13.5 g of barium chloride using its molar mass (208.23 g/mol).
[tex]13.5 g \times \frac{1mol}{208.23g} = 6.5 \times 10^{-2} mol[/tex]
A 427 g sample is 3.17% by mass barium chloride contains 6.5 × 10⁻² moles of BaCl₂.
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