The correct answer is option B, that is, add 1.46 grams of NaCl to 250 milliliters of H₂O.
First there is a need to find the moles of NaCl in 250 ml of 0.10 M NaCl,
Moles of NaCl = molarity × volume = 0.10 M × (250/1000L) = 0.025 mol
The corresponding mass of NaCl is,
Mass of NaCl = moles × molar mass = 0.025 mol × 58.5 g/mol = 1.46 g
Thus, there is a need to dissolve 1.46 grams of NaCl solid into 50 ml of H₂O and dilute to 250 ml.
<span>Aqueous solution is something where water is solvent. When the aqueous solution is saturated in both potassium chlorinate and carbon dioxide gas at 85C, the carbon dioxide bubbles out of the solution. The hydrophobic substances do not get dissolved in the water.</span>
Hello!
If there's an air bubble inside the buret, and the bubble escapes the buret during the titration the initial volume lecture (Vi) would be lower (closer to 0) than the actual one, and the recorded consumed volume (ΔV=Vf-Vi) would be higher than the actual one and thus the calculated concentration of the hydrochloric acid would be higher than the real one.
Have a nice day!
Balanced Equation is
4H2SiCl2+4H2O → H8Si4O4 + 8HCl
