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hjlf
3 years ago
12

For problems 2 & 3, the density of the sample is 0.9977 g/mL and the molar mass of calcium carbonate is 100.0 g/mol. 2. Calc

ulatethemass(ingrams)ofcalciumcarbonatepresentina50.00mLsampleof an aqueous calcium carbonate standard, assuming the standard is known to have a hardness of 75.0 ppm (hardness due to CaCO3).
Chemistry
1 answer:
Alenkasestr [34]3 years ago
4 0
1) Use of density formula to calculate mass of sample, M

D = M / V => M = D* V = 0.9977 g / mL * 50.00 mL = 49.885 g of sample

Realize that the sample is the solution.

2) Use ppm concentration to calculate mass of solute (calcium carbonate)

By definition, 75.0 ppm = 75.0 g of calcium carbonate / 1,000,000 g of solution.

You use that ratio to calculate the mass of calcium carbonate in 49.885 g of solution.

Mass of solute = 49.885 g of solution * [75.0 g of calcium carbonate] / [1,000,000 g of solution] =

Mass of solute =  0.00374 g of calcium carbonate

Answer: 0.00374 g of calcium carbonate
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mixture of N 2 And H2 Gases weighs 13.22 g and occupies a volume of 24.62 L at 300 K and 1.00 atm.Calculate the mass percent of
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<u>Answer:</u> The mass percent of nitrogen gas and hydrogen gas is 91.41 % and 8.59 % respectively.

<u>Explanation:</u>

To calculate the number of moles, we use the equation given by ideal gas equation:

PV = nRT

where,

P = Pressure of the gaseous mixture = 1.00 atm

V = Volume of the gaseous mixture = 24.62 L

n = number of moles of the gaseous mixture = ?

R = Gas constant = 0.0821\text{ L atm }mol^{-1}K^{-1}

T = Temperature of the gaseous mixture = 300 K

Putting values in above equation, we get:

1.00atm\times 24.62L=n_{mix}\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 300K\\\\n_{mix}=\frac{1.00\times 24.62}{0.0821\times 300}=0.9996mol

We are given:

Total mass of the mixture = 13.22 grams

Let the mass of nitrogen gas be 'x' grams and that of hydrogen gas be '(13.22 - x)' grams

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

<u>For nitrogen gas:</u>

Molar mass of nitrogen gas = 28 g/mol

\text{Moles of nitrogen gas}=\frac{x}{28}mol

<u>For hydrogen gas:</u>

Molar mass of hydrogen gas = 2 g/mol

\text{Moles of hydrogen gas}=\frac{(13.22-x)}{2}mol

Equating the moles of the individual gases to the moles of mixture:

0.9996=\frac{x}{28}+\frac{(13.22-x)}{2}\\\\x=12.084g

To calculate the mass percentage of substance in mixture we use the equation:

\text{Mass percent of substance}=\frac{\text{Mass of substance}}{\text{Mass of mixture}}\times 100

Mass of the mixture = 13.22 g

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Mass of nitrogen gas = x = 12.084 g

Putting values in above equation, we get:

\text{Mass percent of nitrogen gas}=\frac{12.084g}{13.22g}\times 100=91.41\%

  • <u>For hydrogen gas:</u>

Mass of hydrogen gas = (13.22 - x) = (13.22 - 12.084) g = 1.136 g

Putting values in above equation, we get:

\text{Mass percent of hydrogen gas}=\frac{1.136g}{13.22g}\times 100=8.59\%

Hence, the mass percent of nitrogen gas and hydrogen gas is 91.41 % and 8.59 % respectively.

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