Question

375 mL of a 0.455 M sodium

chloride solution is diluted with
1.88 L of water.
What is the new concentration in
molarity?
[?]M NaCl

Answer 1

Step-by-step explanation:

Answer : The new concentration will be, 0.0757 M

Solution :

According to the dilution law,

where,

= molarity of NaCl solution = 0.455 M

= volume of NaCl solution = 375 ml = 0.375 L

= new concentration = ?

= volume of new concentration after dilution with water = 0.375 + 1.88 = 2.255 L

Now put all the given values in the above law, we get the new concentration.

Therefore, the new concentration will be, 0.0757 M

Answer 2

0.0757 M NaCl

Step-by-step explanation:

To calculate the new concentration of the sodium chloride (NaCl) solution after dilution, we can use the dilution formula.

Dilution formula

`\boxed{\sf C_1V_1=C_2V_2}`

where:

• C₁ = Initial concentration
• V₁ = Initial volume
• C₂ = Final concentration
• V₂ = Final volume

In this case, we have 375 mL of a 0.455 M sodium chloride solution which is diluted with 1.88 L (1880 mL) of water. The final volume of the solution after dilution is:

`\sf V_2 = V_1 + V_(\sf water) = 375\;mL + 1880\; mL = 2255\; mL`

Therefore:

• C₁ = 0.455 M
• V₁ = 375 mL
• V₂ = 2255 mL

Substitute these values into the formula and solve for C₂:

`\implies \sf 0.455\;M \cdot 375\;mL=C_2 \cdot 2255\;mL`

`\implies \sf C_2=(0.455\;M \cdot 375\;mL)/(2255\;mL)`

`\implies \sf C_2=0.075665188...\; M`

`\implies \sf C_2=0.0757\; M\;(3\;s.f.)`

Therefore, the new concentration of the sodium chloride solution after dilution is 0.0757 M, to three significant figures.