Question

you are given 10.5m hydrochloric acid concentrated solution. if you want to prepare 525 ml of 4.5m dilute solution, then, what volume of concentrated solution and what volume of water do you need? (hint: $m_1v_1 = m_2v_2$ formula). 1) 10.5ml acid, 221.4ml water 2) 300ml acid, 225ml water 3) 225ml acid, 300ml water 4) 525 ml acid, no water

Explanation:

Step1: Identify known values

We know the formula for dilution is \( M_1V_1 = M_2V_2 \), where \( M_1 = 10.5 \, M \) (concentration of concentrated solution), \( M_2 = 4.5 \, M \) (concentration of dilute solution), and \( V_2 = 525 \, \text{mL} \) (volume of dilute solution). We need to find \( V_1 \) (volume of concentrated solution).

Step2: Solve for \( V_1 \)

Rearrange the formula to solve for \( V_1 \): \( V_1=\frac{M_2V_2}{M_1} \)
Substitute the known values: \( V_1=\frac{4.5 \, M\times525 \, \text{mL}}{10.5 \, M} \)
First, calculate the numerator: \( 4.5\times525 = 2362.5 \)
Then divide by the denominator: \( V_1=\frac{2362.5}{10.5}=225 \, \text{mL} \)? Wait, no, wait, let's recalculate. Wait, \( 4.5\times525 = 2362.5 \), \( 2362.5\div10.5 = 225 \)? Wait, no, wait, maybe I made a mistake. Wait, no, the options have 225 and 300. Wait, wait, no, let's check again. Wait, \( M_1 = 10.5 \), \( M_2 = 4.5 \), \( V_2 = 525 \). So \( V_1=\frac{4.5\times525}{10.5}=\frac{2362.5}{10.5}=225 \)? Wait, but option 3 is 225 mL acid, 300 mL water. Wait, the volume of water is \( V_2 - V_1 = 525 - 225 = 300 \, \text{mL} \). Wait, so \( V_1 = 225 \, \text{mL} \) (concentrated acid), and water is \( 525 - 225 = 300 \, \text{mL} \). So the correct option is 3) 225mL acid, 300mL water. Wait, but in my first calculation, I thought I messed up, but let's check again. \( 4.5\times525 = 2362.5 \), divided by 10.5: 10.5×225 = 2362.5, yes. So \( V_1 = 225 \, \text{mL} \). Then volume of water is \( 525 - 225 = 300 \, \text{mL} \). So option 3 is correct.

Answer:

1. 225mL acid, 300mL water