Question

Please help fast!! I need someone to see if my answers are correct! If any are wrong please let me know what the correct answers would be and how to get that answer!

1. how many moles of sodium chloride are in 28 grams od NaCl?

A. .265 mole NaCl
B. .856 mole NaCl
C. .479 mole of NaCl
D. 1.2 mole NaCl <my choice

2. 734 grams of lithium sulfate (Li2SO4) are dissolved to make 2500 mL of solution what is rhe molaratiy?

A. 2.67 M
B. 4.56 M
C. 3.89 M <my choice
D. 1.78 M

3. how many grams of CaCl2 would be dissolved in 3.0 L of a 0.50 M solution of CaCl2?

A. 250 g CaCl2
B. 166.5 g CaCl2
C. 113.65 g CaCl2
D. 98 g CaCl2 <my choice

4. Suppose you had 58.44 g of NaCl and you dissolved it in exactly 2.00 liters. The molarity if the solution would be 0.5 M

true <my choice
false

5. I would need 22g of NaOH to make a 3.0 M solution using 250 mL of solvent.

true <my choice
false

6. Identify the solute:
you have a .0195 M solution made from using 6.5 g of solute and 3 L of solvent. Identify the solute by solving for molar weight.

A. the solute is NaCl because the molar weight is 58.43 g/mol <my choice

B. the solute is H2SO4 because the molar weight is 98.06 g/mol

C. the solute is CaCl2 because the molar weight is 111.11 g/mol

Answer 1

Yeah all of them are right :)

Answer 2

• 1) C. 0.479 mole of NaCl
• 2)Therefore, the correct answer is: A. 2.67 M
• 3) the correct answer is B) 166.5 g CaCl2.
• 4) the molarity of the solution is indeed 0.5 M, making the statement true.
• 5) 22g of NaOH would not be sufficient to make a 3.0 M solution using 250 mL of solvent. The correct answer is false.
• 6) the solute in the .0195 M solution is CaCl2, not NaCl (58.43 g/mol) or H2SO4 (98.06 g/mol).

Step-by-step explanation:

1) To find the number of moles of sodium chloride (NaCl) in 28 grams, you need to use the formula:

moles = mass / molar mass

The molar mass of NaCl is 58.44 g/mol.

Now, let's calculate:

moles = 28 g / 58.44 g/mol

moles = 0.479 mol

Therefore, the correct answer is:

C. 0.479 mole of NaCl

2) To find the molarity of a solution, you need to use the formula:

Molarity = moles of solute / volume of solution (in liters)

First, we need to find the number of moles of lithium sulfate (Li2SO4) in 734 grams.

To do this, we need to calculate the molar mass of Li2SO4:

Molar mass of Li2SO4 = (2 x molar mass of Li) + molar mass of S + (4 x molar mass of O)

The molar mass of Li is 6.94 g/mol, the molar mass of S is 32.07 g/mol, and the molar mass of O is 16.00 g/mol.

Now, let's calculate the molar mass of Li2SO4:

Molar mass of Li2SO4 = (2 x 6.94 g/mol) + 32.07 g/mol + (4 x 16.00 g/mol)

Molar mass of Li2SO4 = 109.94 g/mol

Next, we can calculate the number of moles of Li2SO4:

moles = mass / molar mass

moles = 734 g / 109.94 g/mol

moles = 6.68 mol

Now, we need to convert the volume of the solution from milliliters (mL) to liters (L).

volume = 2500 mL = 2500/1000 L = 2.5 L

Finally, we can calculate the molarity:

Molarity = moles of solute / volume of solution (in liters)

Molarity = 6.68 mol / 2.5 L

Molarity = 2.67 M

Therefore, the correct answer is:

A. 2.67 M

3) To determine the amount of CaCl2 dissolved in 3.0 L of a 0.50 M solution of CaCl2, we can use the formula:

Molarity (M) = moles of solute / volume of solution in liters

First, we need to find the number of moles of CaCl2 in the solution.

1. Molarity is given as 0.50 M, which means there are 0.50 moles of CaCl2 in 1 liter of the solution.

2. We can calculate the moles of CaCl2 in 3.0 L of the solution by multiplying the molarity by the volume:

Moles of CaCl2 = 0.50 M * 3.0 L

= 1.5 moles.

3. Now, we can calculate the mass of CaCl2 using the molar mass of CaCl2, which is the sum of the atomic masses of calcium (Ca) and chlorine (Cl):

Molar mass of CaCl2 = (1 * atomic mass of Ca) + (2 * atomic mass of Cl)

= (1 * 40.08 g/mol) + (2 * 35.45 g/mol)

= 40.08 g/mol + 70.90 g/mol

= 111.98 g/mol.

4. Finally, we can calculate the mass of CaCl2 dissolved in the solution by multiplying the moles of CaCl2 by the molar mass:

Mass of CaCl2 = 1.5 moles * 111.98 g/mol

= 167.97 g.

Therefore, the correct answer is B) 166.5 g CaCl2.

To determine the molarity of a solution, we need to use the formula:

Molarity (M) = moles of solute / volume of solution (in liters)

In this case, we are given the mass of NaCl (58.44 g) and the volume of the solution (2.00 liters).

To find the moles of NaCl, we can use the formula:

moles = mass / molar mass

The molar mass of NaCl is 58.44 g/mol (22.99 g/mol for Na + 35.45 g/mol for Cl).

So, moles of NaCl = 58.44 g / 58.44 g/mol = 1 mol

Now we can substitute the values into the molarity formula:

Molarity (M) = 1 mol / 2.00 L = 0.5 M

Therefore, the molarity of the solution is indeed 0.5 M, making the statement true.

5) we can calculate the mass of NaOH needed using the molar mass of NaOH, which is the sum of the atomic masses of sodium (Na), oxygen (O), and hydrogen (H):

Molar mass of NaOH = (1 * atomic mass of Na) + (1 * atomic mass of O) + (1 * atomic mass of H)

= (1 * 22.99 g/mol) + (1 * 16.00 g/mol) + (1 * 1.01 g/mol)

= 22.99 g/mol + 16.00 g/mol + 1.01 g/mol

= 40.00 g/mol.

Mass of NaOH = 0.75 moles * 40.00 g/mol

= 30.00 g.

Therefore, 22g of NaOH would not be sufficient to make a 3.0 M solution using 250 mL of solvent. The correct answer is false.

6.To identify the solute in a .0195 M solution, we can use the formula:

Molarity (M) = moles of solute / liters of solution

Given that the molarity is .0195 M and the volume of the solution is 3 L, we can rearrange the formula to solve for the moles of solute:

moles of solute = Molarity x volume of solution

moles of solute = .0195 M x 3 L

moles of solute = .0585 moles

To determine the molar weight of the solute, we need to know the mass of the solute. Given that 6.5 g of solute was used, we can use the formula:

moles of solute = mass of solute / molar weight

By rearranging the formula, we can solve for the molar weight:

molar weight = mass of solute / moles of solute

molar weight = 6.5 g / .0585 moles

molar weight ≈ 111.11 g/mol

Therefore, the solute in the .0195 M solution is CaCl2, not NaCl (58.43 g/mol) or H2SO4 (98.06 g/mol).