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3 years ago

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The solubility of in water at a certain temperature is 35.7 g /100. g . Suppose that you have 330.0 g of . What is the minimum v

olume of water you would need to dissolve it all? (Assume that the density of water is 1.00 g/mL.)

1 answer:

juin [17]3 years ago

8 0

The question is incomplete, here is the complete question:

The solubility of substance X in water at a certain temperature is 35.7 g /100. g. Suppose that you have 330.0 g of substance X. What is the minimum volume of water you would need to dissolve it all? (Assume that the density of water is 1.00 g/mL.)

<u>Answer:</u> The minimum volume of water that would be needed is 940.17 mL

<u>Explanation:</u>

We are given:

Solubility of substance X in water = 35.1 g/100 g

This means that 35.1 grams of substance X is dissolved in 100 grams of water

Applying unitary method:

If 35.1 grams of substance X is dissolved in 100 grams of water

So, 330.0 grams of substance X will be dissolved in = $\frac{100}{35.1}\times 330=940.17g$ of water

To calculate the volume of water, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of water = 1 g/mL

Mass of water = 940.17 g

Putting values in above equation, we get:

$1g/mL=\frac{940.17g}{\text{Volume of water}}\\\\\text{Volume of water}=\frac{940.17g}{1g/mL}=940.17mL$

Hence, the minimum volume of water that would be needed is 940.17 mL

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<u>Answer:</u> The empirical and molecular formula for the given organic compound is $CHO_2$ and $C_2H_2O_4$

<u>Explanation:</u>

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=13.00g$

Mass of $H_2O=2.662g$

We know that:

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<u>For calculating the mass of carbon:</u>

In 44 g of carbon dioxide, 12 g of carbon is contained.

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<u>For calculating the mass of hydrogen:</u>

In 18 g of water, 2 g of hydrogen is contained.

So, in 2.662 g of water, $\frac{2}{18}\times 2.662=0.296g$ of hydrogen will be contained.

Mass of oxygen in the compound = (13.30) - (3.54 + 0.296) = 9.464 g

To formulate the empirical formula, we need to follow some steps:

<u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{3.54g}{12g/mole}=0.295moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.296g}{1g/mole}=0.296moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{9.465g}{16g/mole}=0.603moles$

<u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.295 moles.

For Carbon = $\frac{0.295}{0.295}=1$

For Hydrogen = $\frac{0.296}{0.295}=1$

For Oxygen = $\frac{0.603}{0.295}=2.044\approx 2$

<u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : H : O = 1 : 1 : 2

Hence, the empirical formula for the given compound is $CHO_2$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

$n=\frac{\text{Molecular mass}}{\text{Empirical mass}}$

We are given:

Mass of molecular formula = 90.04 g/mol

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Putting values in above equation, we get:

$n=\frac{90.04g/mol}{45g/mol}=2$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(1\times 2)}H_{(1\times 2)}O_{(2\times 2)}=C_2H_2O_4$

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