Respuesta :
Answer: Measure 114 mL of the 1.75 M solution, and dilute it to 2.00 L.
Explanation:
The number of moles in the solution will remain same on dilution and thus according to Molarity Equation:
[tex]M_1V_1=M_2V_2[/tex]
[tex]M_1[/tex]= molarity of first solution
[tex]V_1[/tex]= Volume of first solution
[tex]M_2[/tex]= molarity of second solution
[tex]V_2[/tex]= Volume of second solution
[tex]0.1\times 2.00L=1.75M\times V_2[/tex]
[tex]V_2=0.114L=114ml[/tex]
Thus 114 ml of 1.75 M solution is taken and the volume is diluted to 2.0 L to make 0.100 M solution.
Hello!
A student requires 2.00 L of 0.100 M NH4NO3 from a 1.75 M NH4NO3 stock solution. What is the correct way to get the solution ?
Measure 114 mL of the 1.75 M solution, and dilute it to 1.00 L.
Measure 114 mL of the 1.75 M solution, and dilute it to 2.00 L.
Measure 8.75 mL of the 1.75 M solution, and dilute it to 2.00 L.
Measure 8.75 mL of the 0.100 M solution, and dilute it to 2.00 L.
We have the following data:
M1 (initial molarity) = 0.100 M (or mol/L)
V1 (initial volume) = 2.00 L
M2 (final molarity) = 1.75 M (or mol/L)
V2 (final volume) = ? (in L or mL)
Let's use the formula of dilution and molarity, so we have:
[tex]M_{1} * V_{1} = M_{2} * V_{2}[/tex]
[tex]0.100 * 2.00 = 1.75 * V_{2}[/tex]
[tex]0.2 = 1.75\:V_2[/tex]
[tex]1.75\:V_2 = 0.2[/tex]
[tex]V_2 = \dfrac{0.2}{1.75}[/tex]
[tex]\boxed{\boxed{V_2 \approx 0.114\:L}}\:\:or\:\:\:\boxed{\boxed{V_2 \approx 114\:mL}}\:\:\:\:\:\bf\green{\checkmark}[/tex]
Answer:
Measure 114 mL of the 1.75 M solution, and dilute it to 2.00 L.
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[tex]\bf\green{I\:Hope\:this\:helps,\:greetings ...\:Dexteright02!}\:\:\ddot{\smile}[/tex]
