AP Chemistry - Identifying Unknown Concentration

39mL of a solution of $10^{-4}M$ is mixed with 24mL of an solution of unknown concentration produced a solution with a pH of 7. What is the concentration of the solution?

$1.63\cdot10^{-4}M$

None of these

$3.75\cdot10^{-4}$

$2.5\cdot10^{-4}$

$1.5\cdot10^{-4}$

Explanation

For this question use the following formula:

is the number of acidic hydrogens on the acid, is the molarity of the acid, is the volume of the acid, is the number of basic hydroxides on the base, is the molarity of the base, is the volume of the base

Rearrange the equation for the molarity of the acid:

$\frac{N_B M_B V_B }{N_A V_A}= M_A$

Plug in known values and solve.

$M_A=1.63\cdot10^{-4}M$

A 100mL solution is composed of 25% ethanol by volume and water. What is the mass of the solution?

$\rho_{ethanol}=0.789\frac{g}{mL}$

Explanation

First we determine the mass of the ethanol in solution using its density. Using the percent by volume of ethanol, we know that there are 25mL of ethanol in a 100mL solution. The remaining 75mL are water.

$25\hspace{ 1 mm}mL\hspace{ 1 mm}EtOH\times\frac{0.789\hspace{ 1 mm}g}{1\hspace{ 1 mm}mL}=19.7\hspace{ 1 mm}g\hspace{ 1 mm}EtOH$

Since the density of water is 1g/mL, we know that the mass of 75mL of water is 75g. The total mass is the sum of the ethanol and the water.

$19.7\hspace{ 1 mm}g+75.0\hspace{ 1 mm}g= 94.7\hspace{ 1 mm}g$

What is the concentration if 40 g NaOH is dissolved in 1000 g of water

1 M

1 m

40 M

4 m

4 M

Explanation

first of all, M = molar; m = molal- M = mol solute/ L of solution; m = mol solute/ kg solvent

you have 40 g NaOH * 1 mol/40 g = 1 mol

1000 g of water is equivalent to 1 L

1 mol/L = 1M

80.0g NaOH is put into 50000 mL water. What is the molarity of the resulting solution?

0.0016 M

0.00004 M

1.6 M

0.625 M

0.04 M

Explanation

Molarity = mol solute / L soution

mol solute = 80 g NaOH * 1 mol / 40 g = 2 mol

L solution = 50000 mL water * 1 L/1000 mL = 50 L

2 mol / 50 L =

30mL of a solution of of unknown concentration mixed with 19mL of $10^{-3}M$ produced a solution with a pH of 7. What is the concentration of the solution?

$6.3\cdot 10^{-4}$

None of these

$6.3\cdot 10^{-3}$

Explanation

For this question use the following formula:

Rearrange the equation for the molarity of the base:

$\frac{N_A M_A V_A }{N_B V_B}= M_B$

Plug in known values and solve.

$M_B=6.3\cdot10^{-4}M$

Which solution has the highest molarity?

$\486.5g\ \text{of}\ Pb(NO_3)_2\ \text{in}\ 0.75L\ H_2O$

$\174g\ \text{of}\ NaCl\ \text{in}\ 2.0L\ H_2O$

$\240g\ \text{of}\ MgSO_4\ \text{in}\ 3.0L\ H_2O$

$\85.0g\ \text{of}\ NaNO_3\ \text{in}\ 2.0L\ H_2O$

$\138.0g\ \text{of}\ CH_3CH_2OH\ \text{in}\ 3.0L\ H_2O$

Explanation

This question requires us to calculate molarity for each answer choice. It is important to add everything correctly and be careful with more complex compounds.

Molarity is simply moles of solute over liters of solution. The correct answer, after trying each, is the answer with lead (II) nitrate, as it gets us a molarity of 2.

$Pb(NO_3)_2:\ \frac{486.5g*\frac{1mol}{331g}}{0.75L}=1.96M$

$NaCl:\ \frac{174g*\frac{1mol}{58.5g}}{2.0L}=1.49M$

$MgSO_4:\ \frac{240g*\frac{1mol}{120.3g}}{3.0L}=0.67M$

$NaNO_3:\ \frac{85g*\frac{1mol}{85g}}{2.0L}=0.50M$

$CH_3CH_2OH:\ \frac{138g*\frac{1mol}{46g}}{3.0L}=1.0M$

Find the mass of in 10L of water if it is a 2m solution.

Explanation

Molality is grams of solute per kilogram of solvent.

$m=\frac{mol}{kg}\rightarrow mol=(m)(kg)$

Water has a density of one gram per mililiter, so one liter of water equal to one kilogram. If we have a 2m solution, that means we have two moles of per kilogram of water.

$10L*\frac{1kg}{1L}=10kg$

has a molecular weight of $\small 56\frac{g}{mol }$ .

$20mol*\frac{56g}{1mol}=1120g\ KOH$

This gives us of .

How many liters of 10M HCl are needed to make 4L of 0.4M solution?

Explanation

A simple calculation can be done to perform any solution dilution problem. We know our equation $molarity = \frac{moles}{volume}$ .

We can rewrite this as .

Using this formula, we take the old solution and set it equal to the new solution.

$V = \frac{1.6mol}{10M}=0.16L$

We need 0.16 liters of our 10 molar solution.

29mL of a solution of $10^{-4}M$ is mixed with 65mL of an solution of unknown concentration produced a solution with a pH of 7. What is the concentration of the solution?

$4.5\cdot10^{-5}M$