Question

1000.0 ml of water is added to 300.0 ml of 5.00 m h₂so₄. determine the new molarity. m₁ = 5.00 mol/l, v₁ = 300.0 ml, v_{h₂o} = 1000.0 ml, v_f =? ml, m_f =? mol/l, v_f = v₁+v_{h₂o}=300.0 ml + 1000.0 ml = 1300.0 ml, m₁v₁ = m_fv_f, m_f = \frac{m₁v₁}{v_f}, m_f = \frac{(5.00 mol/l)(300.0 ml)}{1300.0 ml}=1.15 mol/l. notice that in this case, we didnt have to convert the volumes to liters because they cancel out anyways! multiple - choice question. calculate the volume of 12m koh needed to make 3600. ml of 2m koh.

Explanation:

Step1: Determine final volume

$V_f=V_1 + V_{H_2O}=300.0\ mL+1000.0\ mL = 1300.0\ mL$

Step2: Use dilution formula

We know $M_iV_i = M_fV_f$, so $M_f=\frac{M_iV_i}{V_f}$. Given $M_i = 5.00\ mol/L$, $V_i=300.0\ mL$ and $V_f = 1300.0\ mL$. Then $M_f=\frac{(5.00\ mol/L)\times(300.0\ mL)}{1300.0\ mL}=1.15\ mol/L$

For the multiple - choice question on the right:
We use the dilution formula $M_1V_1 = M_2V_2$. Here $M_1 = 12\ M$, $M_2=2\ M$, $V_2 = 3600\ mL$. We need to find $V_1$.
From $M_1V_1 = M_2V_2$, we have $V_1=\frac{M_2V_2}{M_1}=\frac{(2\ M)\times(3600\ mL)}{12\ M}=600\ mL$

Answer:

The new molarity in the first part is $1.15\ mol/L$.
For the multiple - choice question: 600 mL (corresponding to the correct option which is not labeled but should be the one with 600 mL value)