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4 years ago

36.8 g of CuSO4(s) was added to water to prepare a 2.00 L solution. What is the

Chemistry

1 answer:

Yuki888 [10]4 years ago

5 0

Answer:

Option D. 0.115 M

Explanation:

The following data were obtained from the question:

Mass of CuSO4 = 36.8 g

Volume of solution = 2 L

Molar mass of CuSO4 = 159.62 g/mol

Molarity of CuSO4 =..?

Next, we shall determine the number of mole in 36.8 g of CuSO4.

This can be obtained as shown below:

Mass of CuSO4 = 36.8 g

Molar mass of CuSO4 = 159.62 g/mol

Mole of CuSO4 =.?

Mole = mass /Molar mass

Mole of CuSO4 = 36.8 / 159.62

Mole of CuSO4 = 0.23 mole

Finally, we shall determine the molarity of the CuSO4 solution as illustrated below:

Mole of CuSO4 = 0.23 mole

Volume of solution = 2 L

Molarity of CuSO4 =..?

Molarity = mole /Volume

Molarity of CuSO4 = 0.23 / 2

Molarity of CuSO4 = 0.115 M

Therefore, the molarity of the CuSO4 solution is 0.115 M.

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A student placed 10.5 g of glucose (C6H12O6) in a volumetric fla. heggsk, added enough water to dissolve the glucose by swirling

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Answer: The mass of glucose in final solution is 0.420 grams

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Putting values in equation 1, we get:

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$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the molarity and volume of the concentrated glucose solution

$M_2\text{ and }V_2$ are the molarity and volume of diluted glucose solution

We are given:

$M_1=0.583M\\V_1=20.0mL\\M_2=?M\\V_2=0.5L=500mL$

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$0.583\times 20=M_2\times 500\\\\M_2=\frac{0.583\times 20}{500}=0.0233M$

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Putting values in equation 1, we get:

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3 years ago

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tatuchka [14]

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Answer:

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Explanation:

Given data:

Percentage of solution = 32.9

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Formula:

% = [grams of solute / grams of solution] × 100

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