Answer to Question #285388 in Organic Chemistry for buddy

Question

1. What is the concentration, in g/mL, of a solution made by dissolving 75.0 g of phosphoric acid in 50.0 mL of solution?
2. What mass of NaCl is needed to produce 500 mL of a 2.0 mol/L solution?
3. What is the molarity of a solution that contains 0.350 g of barium fluoride, BaF₂, in 0.50 L of solution?
4. What is the molar concentration of a solution containing 10.0 g of NaOH in 250 mL of solution?
5. What mass of silver nitrate, AgNO₃, is needed to make 0.50 L of a 0.10 mol/L solution?
6. What volume of 1.0 mol/L solution of CuS can be made from 23.9 g of CuS?
7. A 100 mL sample of a 6.0 mol/L solution of H₂SO₄ is diluted to a final volume of 500 mL. What is the concentration of the diluted solution?
8. What volume of a 6.0 M NaOH solution is needed to produce 3.0 L of a 0.1 M solution?
9. Explain the steps you would follow to make 500 mL of a 0.5 mol/L solution of sodium chloride (NaCl) in the lab.

Answer 1

The concentration of the solution can be calculated using the following formula:

c = m / V

where c - concentration, m - mass, V - volume of the solution

From here:

c(phosphoric acid) = m(phosphoric acid) / V(phosphoric acid) = 75.0 g / 50.0 ml = 1.50 g/ml

Answer: 1.50 g/mol

1)Convert mL to L:

500mL=0.5L

2) find number of moles NaCl in our solution:

n(NaCl)=C(NaCl)*V(NaCl)=2.0mol/L * 0.5L=1.0 mol

3) find mass of NaCl in 1 mol:

m(NaCl)=n(NaCl)*Mr(NaCl)=1.0mol*58.44 g/mol=58.4 g.

answer:

we need 58.4 g of NaCl.

What is the molarity of a solution that contains 0.350 g of barium fluoride, BaF2, in 0.50 L of solution? Answer: The molarity of a solution is 0.004 M.

What is the molar concentration of a solution containing 10.0 g of NaOH in 250 mL of solution? The number of moles of NaOH in the solution is 10g/40g/mole=0.25moles.

For answering this question you need to calculate amount of mol in final mixture.

0.10 mol per& 1 L

x in& 0.50 L

x=0.05 mol

Now you can calculate mass of silver nitrate, AgNO3:

n= m/Mw

m=n×Mw& ( Mw of AgNO3 = 169.87)

m=169.87 × 0.05= 8.4935

The molar concentration of the solution can be found for the formula

C = n / V(solution);

The amount of substance can also be found for the formula:

n = m / M

For CuS molar mass will be:

M(CuS) = 64 + 32 =96 g/mol;

n(CuS) = 29.3 g / 96 g/mol = 0.305 moles;

So the volume of solution will be

V(solution) = n /C;

V(solution) = 0.305 moles / 1 mol/L = 0.305 L

So the volume of solution is 0.305 L or 305 ml.

Answer: 0.05 L of 6.0 M NaOH solution is needed to produce 3.0 L of 0.1 m solution.