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VikaD [51]
4 years ago
14

A certain molecular compound M has a solubility in acetone of 0.737g/mL at 10.°CCalculate the mass of M that's dissolved in 9.0L

of a saturated solution of M in acetone at this temperature.Be sure your answer has the correct unit symbol and number of significant digits.
Chemistry
1 answer:
Paha777 [63]4 years ago
5 0

Answer:

6633 grams of M are dissolved in 9L

Explanation:

If the compound M has a solubility in acetone of 0.737 g/mL, means that in 1 mLof solution, 0.737 grams of solute is dissolved.

Let's make a rule of three:

9 L = 9000 mL

In 1 mL ___ 0.737 g of M is dissolved

In 9000 mL ___ (9000 . 0.737) /1 = 6633 grams

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Determine the molarity of a 6.0 mole% sulfuric acid solution with SG-a 1.07 Note: Atomic Weight: S (32), O 16); H (O)
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The molarity of a 6.0 mole% sulfuric acid solution is 2.8157 Molar.

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Suppose there are 100 moles in solution:

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Mass of water = 94 mol × 18 g/mol = 1692 g

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Density of solution = D

S.G=\frac{D}{d_w}

d_w = density of water = 1 g/mL

D=S.G\times d_w=1.07\times 1 g/mL=1.07 g/mL

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Volume = \frac{Mass}{Density}

=\frac{2280 g}{1.07 g/mL}=2130.84 mL=2.13084 L

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Molarity = \frac{n}{V(L)}

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V = volume of the solution in L

here we have ,n = 6 moles of sulfuric acid

V = 2.13084 L

So, the molarity of the solution is :

Molarity=\frac{6 mol}{2.13084 L}=2.8157 mol/L

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