Solubility and Ions - MCAT Physical

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Question

A chemist has 300mL of a 0.75M solution and needs to dilute it to 0.35M. How much solvent should be added?

Answer

Use the equation for molarity, $\frac{# moles}{# liters} = M$ , first to figure out the number of moles of solute present.

$\frac{moles}{0.300 liters} = 0.75 M$

Solving this gives 0.225 moles of solute. Using the same equation again with the final molarity will give the final volume.

$\frac{0.225mol}{x\ L}=0.35M\rightarrow x=0.643L$

Finally, the amount of solvent added is the difference between initial and final total volumes.

0.643L – 0.300L = 0.342L added

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Question

What is the molarity of a 1L solution composed of water and 300g of sodium iodide?

Answer

Sodium iodide is given by the formula NaI, and has a molecular weight of 150g/mol. Molarity is found by dividing the moles of solute by liters of solvent.

$M=\frac{mol\ solute}{L\ solvent}=\frac{mol\ NaI}{L\ water}$

We find the moles of sodium iodide by using the mass (300g) and molecular weight.

$300g*\frac{1mol}{150g}=2mol NaI$

We know our volume is 1L, so now we can solve for the molarity.

$Molarity = \frac{2mol}{1L} = 2M$

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Question

Which of the following aqueous solutions is the most concentrated?

Answer

20 grams of ammonia is approximately 1.17 moles of ammonia, so both the molarity and molality of this solution would not approach 2.

$20g\ NH_3*\frac{1mol\ NH_3}{17g\ NH_3}=1.17mol\ NH_3$

$M=\frac{1.17mol}{1L}<2$

$m=\frac{1.17mol}{1kg}<2$

Now, we need to determine whether molarity or molality is more concentrated. It helps to remember for the MCAT that 1 liter of water is equal to 1 kilogram of water. Molality uses the denomination of kilograms of water, and molarity mixes the solute until liters of solution are created. Since the mixed solution will incorporate the solute, it will require less than 1 kilogram of water. As a result, 2M is more concentrated than 2m.

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Question

What is the final concentration of 100mL of 6M HCl solution when it is diluted with 500mL of water?

Answer

To solve this question we must first determine the number of moles of HCl in the initial solution. Converting 100mL into 0.1L, there is initially 0.6mol of HCl in the solution.

$6\frac{mol}{L}100mL\frac{1L}{1000mL}=0.6mol$

When we dilute the solution with 500mL of water, the final volume of solution is 600mL.

$100mL\ solution+500mL\ H_2O=600mL\ total$

To find molarity we can divide number of moles of solute by the total volume of solution.

$600mL*\frac{1L}{1000mL}=0.6L$

$\frac{0.6mol\ HCl}{0.6L}=1M$

The final concentration is 1M.

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Question

What is the molality of a solution when 300mmol of HBr is added to 60g of ether?

Answer

Molality is the number of moles of solute divided by kilograms of solvent. Because we are given the solute in mmol and the solvent in grams, we must convert them to moles and kilograms, respectively.

300 mmol is 0.3 moles. Additionally 60 grams is 6 x 10-2 kg.

$300mmol\frac{1mol}{1000mmol}=0.30mol$

$60g\frac{1kg}{1000g}=0.06kg$

Now we can solve for the molality.

$molality = \frac{0.30mol}{0.06kg}= 5 m$

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Question

A lab technician prepares a $\small 1.5M$ aqueous solution of copper (II) sulfate. The solution has a density of $\small 1300\frac{g}{L}$ . What is the ratio of moles of copper (II) sulfate to moles of water in this solution?

Answer

First, find the mass of copper (II) sulfate in one liter of solution, using the molar mass of copper (II) sulfate.

$\frac{1.5 mol CuSO_{4}}{1 L}\frac{159.6 g CuSO_{4}}{1 mol CuSO_{4}}=\frac{239.4g CuSO{4}}{1L}$

We now have the density of the copper (II) sulfate in the solution. Using the total density of the solution, we can calculate the contribution from water to this density.

$\frac{1300g\ total}{1L}-\frac{239.4g\ CuSO_4}{1L}=\frac{1060.6g\ H_2O}{1L}$

Now that we know the mass of water per liter of solution, we need to convert to moles.

$1060.6g H_{2}O\frac{1mol H_{2}O}{18.0g H_{2}O}=58.9mol H_{2}O$

We now have the moles of copper (II) sulfate per liter and the moles of water per liter, allowing us to find the molar ratio in the solution.

$\frac{1.5molesCuSO_{4}}{58.9molesH_{2}O}=0.025$

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Question

Which of the following is true of a $\small 1L$ sample of $\small 4m\ NaCl$ solution?

Answer

Concentration can be measured in several different ways. Molality is moles of solute per kilogram of solvent. Molarity is moles of solute per liter of solution. Normality is equivalents of protons per liter of solution.

$m=\frac{mol}{kg};\ M=\frac{mol}{L};\ N=\frac{mol\ H^+}{L}$

In our question, we are given the molality of the solution.

$4m=\frac{4mol\ NaCl}{1kg\ H_2O}$

When using molality, it is important to realize that the solute occupies volume. One kilogram of water is equal to one liter of water, but there is an additional volume of four moles of solid sodium chloride that is added to the solution. A one-liter sample of this solution will, thus, contain less than one liter of water.

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Question

What is the mole fraction of sucrose ( $C_{12}H_{22}O_{11}$ ) in a $\small 0.5\frac{mol}{kg}$ solution?

Answer

Recall that molality is defined as:

$m = \frac{\text{mol solute}}{\text{kg solvent}}$

In this problem there is one half mole of sucrose per one kilogram of water.

Our goal is to calculate the mole fraction:

$X=\frac{\text{mol solute}}{\text{mol total}}$

$X=\frac{mol C_{12}H_{22}O_{11}}{mol C_{12}H_{22}O_{11}+ mol H_2O}$

We need to calculate the number of moles of water in one kilogram of solution.

$1kg H_2O (\frac{1000g}{1kg})(\frac{1molH_2O}{18.0gH_2O})=55.5 molH_2O$

We now know that, in one kilogram of solution, there will be $\small 0.5mol$ of sucrose and $\small 55.5mol$ of water. Use these values in the mole fraction equation to solve for the fraction of sucrose.

$X=\frac{0.5molC_{12}H_{22}O_{11}}{0.5molC_{12}H_{22}O_{11}+55.5molH_2O}$

$X=0.0089=8.9*10^{-3}$

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Question

If 0.02L of a 0.7N NaCl solution is diluted to 1L, what is the new concentration?

Answer

In this particular scenario, molarity can be used interchangeably with normality. The product of the initial normality and initial volume will be equal to the product of the final normality and final volume.

Our initial normality is 0.7N and our initial volume is 0.02L. We also know our final volume is 1L.

Solve for the final concentration.

$N_2=\frac{(0.7N)(0.02L)}{1L}=0.014N$

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Question

What is the Vant Hoff factor of the molecule Li3PO4?

Answer

The Vant Hoff factor indicates how many particles a solid produces when dissolved in solution. When Li3PO4dissolves in solution, there are three Li+ molecules and one molecule of PO4-3. The Vant Hoff factor is equal to the sum of molecules: $3Li^++1PO_4^{-3}=4\ total\ ions$ .

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Question

Which of the following solutions is NOT a good electrolyte?

Answer

Electrolyte solutions are formed when a compound creates ions once in solution. Carbon dioxide will not create ions in solution, so it is not a good electrolyte.

$NaCl(aq)\rightarrow Na^++Cl^-$

$CH_3COOH(aq)\rightarrow CH_3COO^-+H^+$

(This is the acid dissociation for acetic acid.)

$NaOH(aq)\rightarrow Na^++OH^-$

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Question

A U-shaped tube is filled with water and then split into two sections by a membrane at the lowest part of the tube. The membrane is permeable to water, but is impermeable to ions. 50g of salt are added to the left side of the tube and allowed to enter solution.

Which of the following would not result following the addition of salt?

Answer

Osmotic pressure is defined as the tendency for water to diffuse into a solution via osmosis. Since the membrane is not permeable to ions, the salt ions are unable to cross the membrane and make the concentrations equal on both sides of the tube. Instead, water will flow to the side that has salt until the pressure in the tube with salt equals the forces of entropy. This will result in a higher level of water on the side of the tube with salt in it.

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Question

All of the following are true regarding a solution except __________.

Answer

Solutions can occur in all three phases of matter. They can also occur between two compounds in a single phase. For example, brass is a solution of two metals: zinc and copper.

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Question

Hydrochloric acid and sodium sulfite react in aqueous solution via the process:

$2HCl_{(aq)}+Na_{2}SO_{3}_{(aq)}\rightarrow 2 NaCl_{(aq)}+H_{2}O_{(l)}+SO_{2(g)}$

Which of the following correctly expresses the net ionic equation for this reaction?

Answer

Start by writing the total ionic equation, using the following steps.

1. Break all soluble strong electrolytes, denoted by "(aq)," into their ions, indicating the number and charge of each ion.

2. Bring along unchanged any compounds denoted as "(s)," "(l)," or "(g)."

$2HCl_{(aq)}+Na_{2}SO_{3}_{(aq)}\rightarrow 2 NaCl_{(aq)}+H_{2}O_{(l)}+SO_{2(g)}$

$2H^{+}_{(aq)}+2Cl^{-}_{(aq)}+2Na^{+}_{(aq)}+SO^{-2}_{3}_{(aq)}\rightarrow 2 Na^{+}_{(aq)}+2Cl^{-}_{(aq)}+H_{2}O_{(l)}+SO_{2(g)}$

Remove any ions appearing on both sides of the reaction (any "spectator" ions) to get the net ionic equation.

Spectator ions: $2Na^+,\ 2Cl^-$

$2H^{+}_{(aq)}+SO^{-2}_{3}_{(aq)}\rightarrow H_{2}O_{(l)}+SO_{2(g)}$

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Question

The Haber-Bosch process, or simply the Haber process, is a common industrial reaction that generates ammonia from nitrogen and hydrogen gas. A worker in a company generates ammonia from the Haber process. He then dissociates the gaseous ammonia in water to produce an aqueous solution. Since ammonia is a base, it will accept a proton from water, generating and ammonium ion products. The two reactions involved are:

$N_2_(g_)+3H_2_(g_)\rightarrow2NH_3_(g_);; ;;;;;;;;;;;;;;;;;;;;;;;;;(1)$

$NH_3_(g_)+H_2O_(l_)\rightarrow NH_4^+_(aq_)+OH^-_(aq_);;;;;(2)$

The ammonium ion generated from the dissociation of ammonia is __________.

Answer

Remember that there are two types of ions: cations and anions. Cations are produced when a atom of a molecule loses an electron. A neutral atom will contain no charge; however, when the atom loses an electron the atom becomes positively charged because it will contain more protons than electrons. In reaction 2, the nitrogen in ammonia loses an electron (because it shares the electron with the new hydrogen atom), which produces a positively charged ammonium ion. The ammonium ion is a cation because of this positive charge.

Anions are formed when an atom gains an electron, which makes the overall charge of an atom negative. The hydroxide ions from reaction 2 are anions.

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Question

Adding ammonia to a solution containing a copper hydroxide precipitate has the effect of dissolving the precipitate because __________.

Answer

The copper ion is able to form four covalent bonds. This can be explained with electron configurations.

The copper atom configuration: $Cu = [Ar]4s^{1}3d^{10}$

The copper ion configuration: $Cu^{2+} = [Ar]3d^{9}$

The copper ion can accept electrons that fill empty and levels. In this example, the chemical equation is as follows:

$Cu^{2+}_{(aq)}+ 4NH_{3}_{(aq)}\ \leftrightharpoons\ Cu(NH_{3})_{4}^{2+}_{(aq)}$

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Question

${Fe(H_{2}O)_{6}}^{3+}$ and ${CuCl_{4}}^{2-}$ are alike in that __________.

Answer

Complex ions typically form when coordinate covalent bonds are formed between ligands, such as water or chloride, and metal ions that have lost electrons and have available orbitals. While one answer choice, "Both compounds are able to form coordinate covalent bonds with a metal ion," may seem correct, this happens within the compound. The compounds themselves are ions that will form ionic compounds.

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Question

Which is an inaccurate depiction of the central iron atom in a hemoglobin molecule?

Answer

The central iron in hemoglobin is the middle of a complex compound and is involved in six covalent bonds (coordination number of 6). It is bonded to five nitrogen atoms, one of which is part of an amino acid in the globin protein. This leaves one remaining bond, which can be with water, oxygen, or carbon monoxide (which forms a very stable bond, making it toxic).

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Question

Electrolytes play a big role in maintaining blood pressure. Loss of electrolytes often leads to a drop in blood pressure. What can a doctor prescribe to a patient who has very low blood pressure?

Answer

The question states that the patient has a low blood pressure, suggesting that the patient has a low concentration of electrolytes in his blood. To counter this, the doctor must prescribe a solution that will increase the concentration of electrolytes in blood. The best solution would be to increase the consumption of salt, or sodium chloride. Sodium chloride is a strong electrolyte and will dissociate into sodium ions and chlorine ions. This will increase the blood pressure of the patient and will remedy the problem.

Spending an hour a day in a hot sauna will cause excessive sweating. Remember that salts are excreted through sweat; therefore, the electrolyte concentration, and the patient's blood pressure, will decrease further if the patient spends time in a sauna. Although lithium chloride is a strong electrolyte (like sodium chloride), the doctor will not prescribe lithium chloride because it is a toxic substance.

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Question

Electronegativity is an important concept in physical chemistry, and often used to help quantify the dipole moment of polar compounds. Polar compounds are different from those compounds that are purely nonpolar or purely ionic. An example can be seen by contrasting sodium chloride, NaCl, with an organic molecule, R-C-OH. The former is purely ionic, and the latter is polar covalent.

When comparing more than one polar covalent molecule, we use the dipole moment value to help us determine relative strength of polarity. Dipole moment, however, is dependent on the electronegativity of the atoms making up the bond. Electronegativity is a property inherent to the atom in question, whereas dipole moment is a property of the bond between them.

For example, oxygen has an electronegativity of 3.44, and hydrogen of 2.20. In other words, oxygen more strongly attracts electrons when in a bond with hydrogen. This leads to the O-H bond having a dipole moment.

When all the dipole moments of polar bonds in a molecule are summed, the molecular dipole moment results, as per the following equation.

Dipole moment = charge * separation distance

Given the polarity of O-H bonds in water described in the passage, what is the Ksp expression for the following reaction, assuming it takes place in an aqueous environment?

$CaSO_{4} (s) \rightarrow Ca^{2+} (aq) + (SO_{4}})^{2-} (aq)$

Answer

The Ksp is a standard equilibrium constant for the dissolution reaction of a solute. As such, it follows the standard formula for equilibrium constants, and excludes pure liquids and solids (like water and calcium sulfate!).

$K_{eq}=K_{sp}=\frac{[Products]}{[Reactants]}$

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