Question

fill in the missing words
a) one of the ways to make 250 ml of a 0.077 m solution of a sodium hydroxide, is to mix \boxed{} ml of a 1.2 m solution of sodium hydroxide and \boxed{} ml of water.
b) one of the ways to make 1.0 l of a 0.55 m solution of magnesium sulfate, is to mix \boxed{} ml of a 2.5 m solution of magnesium sulfate and add enough water to make 1.0 l of the solution.

Explanation:

Part (a)

Step1: Use dilution formula \( M_1V_1 = M_2V_2 \)

We know \( M_1 = 1.2 \, M \), \( M_2 = 0.077 \, M \), \( V_2 = 250 \, mL \). Solve for \( V_1 \):
\( V_1=\frac{M_2V_2}{M_1}=\frac{0.077 \times 250}{1.2} \)
Calculate numerator: \( 0.077 \times 250 = 19.25 \)
Then \( V_1=\frac{19.25}{1.2}\approx16.04 \, mL \)

Step2: Find volume of water

Total volume is \( 250 \, mL \), so water volume \( = 250 - 16.04 = 233.96 \, mL \)

Part (b)

Step1: Use dilution formula \( M_1V_1 = M_2V_2 \)

\( M_1 = 2.5 \, M \), \( M_2 = 0.55 \, M \), \( V_2 = 1.0 \, L = 1000 \, mL \). Solve for \( V_1 \):
\( V_1=\frac{M_2V_2}{M_1}=\frac{0.55 \times 1000}{2.5} \)
Calculate: \( \frac{550}{2.5}=220 \, mL \)

Answer:

s:
a) Volume of 1.2 M NaOH: \(\approx 16.0 \, mL\) (or 16.04 mL), Volume of water: \(\approx 234 \, mL\) (or 233.96 mL)
b) Volume of 2.5 M \( MgSO_4 \): \( 220 \, mL \)

(Note: For part (a), rounding may vary; 16.0 mL and 234 mL are approximate. For part (b), the calculation gives exactly 220 mL.)