How many milliliters of 0.130 M HCl are needed to completely neutralize 41.0 mL of 0.107 M Ba(OH)2 solution?

54 mL Ba(OH)2x(0.101 mol Ba(OH)2/1000 mL) x (2 mol OH-/ 1 mol Ba(OH)2 )  = 0.0109 mol OH-

0.0109 mol OH-x (1mol HCl/ 1 mol OH- ) = 0.0109 mol HCl

0.109 mol HCl/(0.130 mol/L HCl) =  0.0839 L HCl

0.0839 L HCl * 1000mL = 83.9 mL of 0.130 M HCl

hey there!

boiling point is the temperature at which the vapor pressure of the liquid equals atmospheric pressure. the vapor pressure of solvent molecules is decreased when a solute is added, so a higher temperature is required to increase the number of solvent molecules in the gas phase above the liquid.

at the freezing point, the vapor pressures of the solid and liquid are equal; a lower temperature is needed to reduce the number of solvent particles above the liquid.

hope this you

tobey