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natita [175]
4 years ago
9

How many grams of kcl are present in 85.0 ml of 2.10 m kcl? how many grams of kcl are present in 85.0 ml of 2.10 m kcl? 3.02 133

13.3 6.34 none of the above?
Chemistry
1 answer:
USPshnik [31]4 years ago
3 0
Molarity (M) = moles of the solute (mol) / Volume of the solution (L)

The molarity of the given KCl solution = 2.10 M
Volume of the given KCl solution = 85.0 mL

Hence,
         2.10 M = Moles of KCl / 85.0 x 10⁻³ L
Moles of KCl = 2.10 M x 85.0 x 10⁻³ L
                     = 0.1785 mol

Moles (mol) = Mass (g) / Molar mass (g mol⁻¹)

Molar mass of KCl = 74.56 g mol⁻¹

Hence, 
  0.1785 mol = Mass of KCl / 74.56 g mol⁻¹
Mass of KCl = 0.1785 mol x 74.56 g mol⁻¹
                    = 13.30 g

Hence, mass of given KCl in 85.0 mL is 13.30 g.
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<u>Explanation:</u>

The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}} ......(1)

We are given:

Given mass of Na = 17.25 g

Molar mass of Na = 23 g/mol

Putting values in equation 1, we get:

\text{Moles of Na}=\frac{17.25g}{23g/mol}=0.75mol

For the given chemical reaction:

2Na+Cl_2\rightarrow 2NaCl

By stoichiometry of the reaction:

2 moles of Na produces 2 moles of NaCl

So, 0.75 moles of Na will produce = \frac{2}{2}\times 0.75=0.75mol of NaCl

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Putting values in above equation, we get:

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The percent yield of a reaction is calculated by using an equation:

\% \text{yield}=\frac{\text{Actual value}}{\text{Theoretical value}}\times 100 ......(1)

Given values:

Percent yield of the product = 88 %

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Plugging values in equation 1:

88\% =\frac{\text{Actual yield}}{43.83g}\times 100\\\\\text{Actual yield}=\frac{88\times 43.83}{100}=38.57g

Hence, the actual yield of the product is 38.57 g

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