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There are 1.92 × 10^23 atoms Mo in the cylinder.
<em>Step 1</em>. Calculate the <em>mass of the cylinder
</em>
Mass = 22.0 mL × (8.20 g/1 mL) = 180.4 g
<em>Step 2</em>. Calculate the<em> mass of Mo
</em>
Mass of Mo = 180.4 g alloy × (17.0 g Mo/100 g alloy) = 30.67 g Mo
<em>Step 3</em>. Convert <em>grams of Mo</em> to <em>moles of Mo
</em>
Moles of Mo = 30.67 g Mo × (1 mol Mo/95.95 g Mo) = 0.3196 mol Mo
<em>Step 4</em>. Convert <em>moles of M</em>o to <em>atoms of Mo
</em>
Atoms of Mo = 0.3196 mol Mo × (6.022 × 10^2<em>3</em> atoms Mo)/(1 mol Mo)
= 1.92 × 10^23 atoms Mo
The molarity of Barium Hydroxide is 0.289 M.
<u>Explanation:</u>
We have to write the balanced equation as,
Ba(OH)₂ + 2 HNO₃ → Ba(NO₃)₂ + 2 H₂O
We need 2 moles of nitric acid to react with a mole of Barium hydroxide, so we can write the law of volumetric analysis as,
V1M1 = 2 V2M2
Here V1 and M1 are the volume and molarity of nitric acid
V2 and M2 are the volume and molarity of Barium hydroxide.
So the molarity of Ba(OH)₂, can be found as,

= 0.289 M
Answer:
I think the answer is A!!!