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Marysya12 [62]
3 years ago
6

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]3 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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The percentages of KBr and KI  in the sample by mass is 80.68 and 19.32 % respectively.

<h3>What is Molar Mass ?</h3>

Molar mass is defined as the mass contained in 1 mole of sample.

It is given that

the sample has a total mass of 5.76 g contains KBr and KI

and contains 1.79 gm of K is present

what is the percentage of KBr and KI in the sample

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In KBr the mass percentage of K is 39/(39+79.9) = 32.89%

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and let the mass of KI present be y

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x = 5.76-1.1125

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% of KBr = (4.6475/5.76  )*100 = 80.68 %

% of KI = (1.1125/5.76) *100 = 19.32%

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