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Bingel [31]
3 years ago
14

Determine the number of formula units of Cr(ClO2)3 and moles of Oxygen contained in 6.85 moles of Cr(ClO2)3

Chemistry
1 answer:
Advocard [28]3 years ago
4 0

The formula  units  of Cr(ClO2)3  that  are in 6.85  moles of Cr(Clo203  is  <u>4.1237 x10^24  formula units</u>


 calculation

  the formula units are  calculated using of Avogadro's  law  constant

that is   according  to Avogadro's law   1  mole = 6.02 x10 ^23  formula units

                                                             6.85 moles= ?  formula units

by  cross  multiplication.

 = ( 6.85 moles   x 6.02  x10^23  formula units )  / 1 mole  = <u>4.1237  x 10 ^24  formula units</u>



<h3>The moles of  Oxygen  contained  in 6.85  moles of cr(ClO2)3  is    <em><u>41.1  moles</u></em></h3>

<em><u>  </u></em><u><em>calculation</em></u>

<em>moles   of  oxygen = number of oxygen atom in  Cr(ClO2)3  x   6.85(moles of  Cr(ClO2)3</em>

calculate the number of   oxygen atoms that  are in Cr(ClO2)3

that is  2 x 3 = 6 atoms of Oxygen.

moles is therefore =  6  x 6.85 moles= 41.1 moles

<h3></h3>

 

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Answer:

Approximately 0.291\; \rm M (rounded to two significant figures.)

Explanation:

The unit of concentration \rm M is the same as \rm mol \cdot L^{-1} (moles per liter.) On the other hand, the volume of both the \rm NaOH solution and the original \rm HCl solution here are in milliliters. Convert these two volumes to liters:

  • V(\mathrm{NaOH}) = 18.2\; \rm mL = 18.2 \times 10^{-3}\; \rm L = 0.0182\; \rm L.
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Calculate the number of moles of \rm NaOH in that 0.0182\; \rm L of 0.160\; \rm M solution:

\begin{aligned} n(\mathrm{NaOH}) &= c(\mathrm{NaOH})\cdot V(\mathrm{NaOH})\\ &= 0.160\; \rm mol \cdot L^{-1} \times 0.0182\; \rm L \approx 0.00291\; \rm mol\end{aligned}.

\rm HCl reacts with \rm NaOH at a one-to-one ratio:

\rm HCl\; (aq) + NaOH\; (aq) \to NaCl\; (aq) + H_2O\; (l).

Coefficient ratio:

\displaystyle \frac{n(\mathrm{HCl})}{n(\mathrm{NaOH})} = 1.

In other words, one mole of \rm NaOH would neutralize exactly one mole of \rm HCl. In this titration, 0.291\; \rm mol of \rm NaOH\! was required. Therefore, the same amount of \rm HC should be present in the original solution:

\begin{aligned}&n(\text{$\mathrm{HCl}$, original})\\ &= n(\mathrm{NaOH})\cdot \frac{n(\mathrm{HCl})}{n(\mathrm{NaOH})} \\ &\approx 0.00291\; \rm mol \times 1 = 0.00291\; \rm mol\end{aligned}.

Calculate the concentration of the original \rm HCl solution:

\displaystyle c(\text{$\mathrm{HCl}$, original}) = \frac{n(\text{$\mathrm{HCl}$, original})}{V(\text{$\mathrm{HCl}$, original})} \approx \frac{0.00291\; \rm mol}{0.0100\; \rm L} \approx 0.291\; \rm M.

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