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antiseptic1488 [7]
3 years ago
6

If you have 16 g of manganese (II) nitrate tetrahydrate, how much water is required to prepare 0.16 M solution from this amount

of salt? OL 0 0.01 L the task is impossible, as manganese nitrate is insoluble in water O 0.10 L 398 ml
Chemistry
1 answer:
nirvana33 [79]3 years ago
7 0

<u>Answer:</u> The volume of water required is 398 mL

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}

We are given:

Mass of solute (manganese (II) nitrate tetrahydrate) = 16 g

Molar mass of manganese (II) nitrate tetrahydrate = 251 g/mol

Molarity of solution = 0.16 M

Putting values in above equation, we get:

0.16M=\frac{16g\times 1000}{251g/mol\times \text{Volume of solution}}\\\\\text{Volume of solution}=398mL

Hence, the volume of water required is 398 mL

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A 250-mL aqueous solution contains 1.56 10–5 g of methanol and has a density of 1.03 g/mL. What is the concentration in ppm?
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To get the concentration in ppm, divide the given mass of methanol by the mass of the solution. Note that the parts-per million (ppm) is equal to mass of solute in milligram(mg) divided by the mass of solution in kilogram (kg)

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7 0
2 years ago
Read 2 more answers
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