Question

1. how many ml of water are required to dilute 2.00 ml of a 5.00 m nacl solution to a final concentration of 1.00 m?

identify the variables ( c_d = 1.00m ) ( c_c = 5.00m )
( v_d =? ) ( v_c = 2.00ml )
first put in the symbols, then put in the numbers
( v_d=\frac{v_c}{c_d}=\frac{(2.00ml)(m)}{1.00m}=ml )
the ml of water added ( =ml - ml =ml )

1. what is the dilution factor? first put in the symbols, then put in the numbers

dilution ( =\frac{v_c}{}=\frac{ml}{ml}= )
a. 3.000 b. 2.500 c. 10.00 d. 1 e. 2.000
f. 9.000 g. 6.000 h. 8.000 i. 0.02500 j. 47.00
k. 5.00 l. 50.00 m. 25.0 n. 0.02041 o. 97.98
p. 30.62 q. 6.572 r. 0.06124 s. 262.87 t. 0.1800
u. ( 3.000\times10^3 ) v. 7.500 w. 0.6667 x. 0.200 y. 2.67
z ( 2.67\times10^3 ) aa. ( 10^3 ) bb. 1 cc. ( c_c ) dd. ( c_d )
ee. ( v_c ) ff. ( v_d )

Explanation:

Step1: Use the dilution formula \(C_CV_C = C_DV_D\)

The formula for dilution is \(C_CV_C = C_DV_D\), where \(C_C\) is the initial concentration, \(V_C\) is the initial volume, \(C_D\) is the final concentration, and \(V_D\) is the final volume. Rearranging for \(V_D\) gives \(V_D=\frac{C_CV_C}{C_D}\).

Step2: Substitute the values into the formula

We know \(C_C = 5.00M\), \(V_C=2.00mL\), and \(C_D = 1.00M\). Substituting these values into the formula \(V_D=\frac{C_CV_C}{C_D}\), we get \(V_D=\frac{(2.00mL)(5.00M)}{1.00M}\).

Step3: Calculate \(V_D\)

\(V_D=\frac{(2.00mL)(5.00M)}{1.00M}=10.00mL\)

Step4: Calculate the volume of water added

The volume of water added is \(V_D - V_C\). Substituting \(V_D = 10.00mL\) and \(V_C=2.00mL\), we get \(10.00mL - 2.00mL = 8.00mL\)

Step5: Calculate the dilution factor

The dilution factor formula is \(\frac{V_D}{V_C}\) (or \(\frac{C_C}{C_D}\)). Using \(\frac{V_D}{V_C}\), with \(V_D = 10.00mL\) and \(V_C=2.00mL\), the dilution factor is \(\frac{10.00mL}{2.00mL}=5\)

Answer:

1.

• 1: \(C_C\)
• 2: \(5.00\)
• 3: \(10.00\)
• 4: \(10.00\)
• 5: \(2.00\)
• 6: \(8.00\)

2.

• 7: \(10.00\)
• 8: \(5\)
• 9: \(V_D\)
• 10: \(2.00\)