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 $m_iv_i = m_fv_f$. 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.

Explanation:

Step1: Identify given values

$M_i = 1.75\ M$, $M_f=0.100\ M$, $V_f = 2.00\ L$

Step2: Rearrange dilution formula for $V_i$

From $M_iV_i = M_fV_f$, we get $V_i=\frac{M_fV_f}{M_i}$

Step3: Substitute values and calculate

$V_i=\frac{0.100\ M\times2.00\ L}{1.75\ M}=\frac{0.200}{1.75}\ L = 0.114\ L=114\ mL$
We need to measure 114 mL of the 1.75 M solution and dilute it to 2.00 L.

Answer:

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