Question

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?

Use mc021-1.jpg.



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.

Answer 1

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:

`M_(1) * V_(1) = M_(2) * V_(2)`

`0.100 * 2.00 = 1.75 * V_(2)`

`0.2 = 1.75\:V_2`

`1.75\:V_2 = 0.2`

`V_2 = (0.2)/(1.75)`

`\boxed{\boxed{V_2 \approx 0.114\:L}}\:\:or\:\:\:\boxed{\boxed{V_2 \approx 114\:mL}}\:\:\:\:\:\bf\green{\checkmark}`

Answer:

Measure 114 mL of the 1.75 M solution, and dilute it to 2.00 L.

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`\bf\green{I\:Hope\:this\:helps,\:greetings ...\:Dexteright02!}\:\:\ddot{\smile}`

Answer 2

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:

`M_1V_1=M_2V_2`

`M_1`= molarity of first solution

`V_1`= Volume of first solution

`M_2`= molarity of second solution

`V_2`= Volume of second solution

`0.1* 2.00L=1.75M* V_2`

`V_2=0.114L=114ml`

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.