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3 years ago

How many grams of cupric sulfate pentahydrate are needed to prepare 50.00 mL of 0.0800M CuSO4× 5H2O?

Chemistry

1 answer:

shepuryov [24]3 years ago

4 0

Explanation:

Molarity is defined as number of moles per liter of solution.

Mathematically, molarity = $\frac{no. of moles}{Volume (in L) of solution}$

It is given that molarity is 0.0800 M and volume is 50.00 mL or 0.05 L.

molarity = $\frac{no. of moles}{Volume of solution in liter}$

0.0800 M = $\frac{no. of moles}{0.05 L}$

no. of moles = 1.6 mol

Therefore, molar mass of cupric sulfate pentahydrate is 249.68 g/mol. So, calculate the mass as follows.

No. of moles = $\frac{mass in grams}{molar mass}$

mass in grams = $no. of moles \times molar mass of CuSO_{4}.5H_{2}O$

= $1.6 mol \times 249.68 g/mol$

= 399.488 g

Thus, we can conclude that 399.488 g of cupric sulfate pentahydrate are needed to prepare 50.00 mL of 0.0800M CuSO4× 5H2O.

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3 years ago

A tablet of Pain Be Gone Aspirin, which had a mass of 1.213 g, was pulverized and 1.159 g were dissolved in 10.0 mL of ethyl alc

Elan Coil [88]

Answer:

a. Moles of $NaOH$ = 0.001643 moles

b. 0.296 g

c. 0.3098 g

d. Not acceptable

Explanation:

Considering:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

Or,

$Moles =Molarity \times {Volume\ of\ the\ solution}$

Given :

For $NaOH$ :

Molarity = 0.1052 M

Volume = 15.62 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 15.62×10⁻³ L

Thus, moles of $NaOH$ :

$Moles=0.1052 \times {15.62\times 10^{-3}}\ moles$

Moles of $NaOH$ = 0.001643 moles

The reaction of NaOH with the acetylsalicylic acid is in the ratio of 1:1.

Thus, Moles of NaOH = Moles of acetylsalicylic acid = 0.001643 moles

Molar mass of acetylsalicylic acid = 180.16 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

Mass = Moles * Molar mass = 0.001643 moles * 180.16 g/mol = 0.296 g

1.159 g of sample contains 0.296 g of acetylsalicylic acid

1.213 g of sample contains $\frac{0.296}{1.159}\times 1.213$ g of acetylsalicylic acid

Mass of acetylsalicylic acid = 0.3098 g = 309.8 mg

d. Sample contains = 309.8 mg

Manufacturer claiming = 315 mg to 335 mg

Thus , it is not acceptable.

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