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?

Answer 1

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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:

$M_{1} * V_{1} = M_{2} * V_{2}$

$0.100 * 2.00 = 1.75 * V_{2}$

$0.2 = 1.75\:V_2$

$1.75\:V_2 = 0.2$

$V_2 = \dfrac{0.2}{1.75}$

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

Answer:

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

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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

$0.1\times 2.00L=1.75M\times V_2$

$V_2=0.114L=114ml$

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.