Question

a student takes a measured volume of 3.00 m hcl to prepare a 50.0 ml sample of 1.80 m hcl. what volume of 3.00 m hcl did the student use to make the sample? use ( m_i v_i = m_f v_f ).
○ 3.70 ml
○ 16.7 ml
○ 30.0 ml
○ 83.3 ml

Explanation:

Step1: Identify known values

We know that \( M_i = 3.00 \, \text{M} \), \( M_f = 1.80 \, \text{M} \), and \( V_f = 50.0 \, \text{mL} \). We need to find \( V_i \) using the formula \( M_iV_i = M_fV_f \).

Step2: Rearrange the formula to solve for \( V_i \)

From \( M_iV_i = M_fV_f \), we can solve for \( V_i \) by dividing both sides of the equation by \( M_i \). So, \( V_i=\frac{M_fV_f}{M_i} \).

Step3: Substitute the known values into the formula

Substitute \( M_f = 1.80 \, \text{M} \), \( V_f = 50.0 \, \text{mL} \), and \( M_i = 3.00 \, \text{M} \) into the formula: \( V_i=\frac{1.80 \, \text{M} \times 50.0 \, \text{mL}}{3.00 \, \text{M}} \).

Step4: Perform the calculation

First, calculate the numerator: \( 1.80 \times 50.0 = 90.0 \). Then divide by the denominator: \( \frac{90.0}{3.00}=30.0 \, \text{mL} \).

Answer:

30.0 mL