Chemistry, 06.07.2021 05:20 TheOneandOnly003

4.106 Calculate the moles and the mass of solute in each of the following solutions.
(a) 150.0 mL of 0.245 M CaCl2

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Answer from: dianaosorio33895

$\boxed {\sf \bold {0.0368 \ mol \ CaCl_2}}}}$

$\boxed {\sf \bold {4.08 \ g \ CaCl_2}}}}}$

Explanation:

1. Moles of Solute

Molarity is a measure of concentration in moles per liter.

$molarity= \frac {moles \ of \ solute}{liters \ of \ solution}$

In this solution, there are 150.0 milliliters of solution and the molarity is 0.245 M CaCl₂ or 0.245 mol CaCl₂ per liter.

First, convert the milliliters to liters. There are 1000 milliliters in 1 liter.

${150 \ mL * \frac{1 \ L}{1000 \ mL}= \frac{150}{1000} \ L = 0.150 \ L$

Now, substitute the known values (molarity and liters of solution) into the formula. The moles of solution are unknown, so we can use x.

$0.245 \ mol \ CaCl_2 /L= \frac{ x}{0.150 \ L}$

We are solving for x, so we must isolate this variable. It is being divided by 0.150 L. The inverse of divisions is multiplication, so we multiply both sides by 0.150 L.

$0.150 \ L *0.245 \ mol \ CaCl_2 /L= \frac{ x}{0.150 \ L} * 0.150 L$

$0.150 \ L *0.245 \ mol \ CaCl_2 /L=x$

The units of liters cancel.

$0.150 *0.245 \ mol \ CaCl_2 =x$

$0.03675 \ mol \ CaCl_2$

The original measurements have 3 significant figures, so our answer must have the same.

We should round to the ten thousandths place. The 5 to the right of this place tells us to round the 7 up to an 8.

$\bold {0.0368 \ mol \ CaCl_2}$

2. Mass of the Solute

We can convert mass to moles using the molar mass. These values are found on the Periodic Table. They are the same as the atomic masses, but the units are grams per mole (g/mol) instead of atomic mass units.

The solute is calcium chloride: CaCl₂. Look up the molar masses of the individual elements.

Ca: 40.08 g/mol Cl: 35.45 g/mol

Notice that chlorine has a subscript of 2. We must multiply the molar mass by 2.

Cl₂: 35.45 *2= 70.9 g/mol

Add calcium's molar mass.

CaCl₂: 40.08 + 70.9 =110.98 g/mol

Use the molar mass as a ratio.

$\frac {110.98 \ g\ CaCL_2}{ 1 \ mol \ CaCl_2}$

Multiply the moles of calcium chloride we calculated above.

$0.0368 \ mol \ CaCl_2 *\frac {110.98 \ g\ CaCL_2}{ 1 \ mol \ CaCl_2}$

The units of moles of calcium chloride cancel.

$0.0368 *\frac {110.98 \ g\ CaCL_2}{ 1 }$

$4.084064 \ g\ CaCl_2$

Round to 3 significant figures again. For this number, it is the hundredths place. The 4 in the thousandths place tells us to leave the 8.

$\bold {4.08 \ g \ CaCl_2}$

Answer from: Quest

Ithink carbon dioxide, i am not completely sure.tell me if i am wrong.

Answer from: Quest

Note that 2 mol dm⁻³ = 2 mol/(dm³)outline: we're given the concentration of hcl, which is 2 mol dm⁻³ (cubic meters in the denominator can represent volume, so i know this is concentration). we're an amount of calcium carbonate caco₃, which is 5 g. we're presented with a chemical equation which includes caco₃ and hcl. we are told to find the volume of hcl. from what we see so far, this seems to be a stoichiometry problem. we'll have to use the given information about calcium carbonate to find the amount of moles of hcl needed to fully react with calcium carbonate. then we'll have information about the moles of hcl and the concentration of hcl; that will be enough to figure out volume according to the formula concentration = moles / volume.======all the answer choices are not in dm.convert the given concentration so that it uses cm⁻³.we can use conversion factors. we have to cube our conversion factors so that dimensional analysis works.we can convert to meters first, using the fact that 10 dm = 1 m ⇔ 10³ dm³ = 1 m³then we use the fact that 10² cm = 1 m ⇔ (10² cm)³ = (1 m)³ ⇔ 10⁶ cm³ = 1 m³so converting: $\dfrac{\text{2 mol}}{\text{dm}^{3}} \cdot \dfrac{\text{10}^3\text{ dm}^3}{\text{1 m}^3} \cdot \dfrac{\text{1 m}^3}{10^6\text{ cm}^3} \\ \\ = \dfrac{\text{2 mol}}{1} \cdot \dfrac{\text{10}^3}{1} \cdot \dfrac{1}{10^6\text{ cm}^3} \qquad \text{\footnotesize cancelling units} \\ \\ = \dfrac{2\text{ mol}}{10^3 \text{ cm}^3} \\ \\ = 0.002\text{ mol/cm}^3$ now we have to find the moles of hydrochloric acid neededfind the moles of calcium carbonate. we have to find the molar mass of calcium carbonate, caco₃, which can be found by molar mass = (mass ca) + (mass c) + 3(mass o) = 40 + 12 + 3(16) = 100 g/molconvert the given 5 g to moles: mol caco₃ = 5 g caco₃ * 1 mol/(100 g) = 0.05 mol caco₃use stoichiometry (ratios) to figure out the moles of hcl needed. ratio wise, there's 2 hcls for every caco₃, so use dimensional analysis mol hcl = mol caco₃ * (2 mol hcl) / (1 mol caco₃) = 0.05 mol hcl * 2 = 0.1 mol hcl we have 0.1 mol hcl and concentration of hcl (0.002 mol/cm³).we need to find volumesince concentration = moles / volume, rearrange to get volume = moles / concentration volume = 0.1 mol / (0.002 mol/cm³) volume = 50 cm³

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4.106 Calculate the moles and the mass of solute in each of the following solutions.
(a) 150....

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