Question 1Tips1 ptsIn this experiment, you will be preparing a standard solution of potassium hydrogen phthalate (MW: 204.2 g.mol⁻¹).  If you weighed out 3.958 g of potassium hydrogen phthalate to prepare a 250.0 mL solution, what would be your concentration (in M)?  Provide your answer to 3 significant figures. Flag question: Question 2Question 2Tips2 ptsThe titration between potassium hydrogen phthalate and sodium hydroxide is a 1:1 stoichiometry.  If the mean titration volume (titre) of sodium hydroxide to reach the endpoint is 29.5 mL against 25 mL of a 7.6 mM standard solution of potassium hydrogen phthalate, what is the concentration of sodium hydroxide (in mM)? Provide your answer to 3 significant figures.Note: this is a hypothetical scenario and the concentrations may not match the previous question or the values used in your experiment. Flag question: Question 3Question 3Tips3 ptsIn this experiment, you will take a 25.00 mL aliquot of vinegar and dilute it to 250.0 mL.  You will then take a 25.00 mL aliquot from this diluted vinegar solution and titrate it against the standardised sodium hydroxide. The titration between acetic acid and sodium hydroxide is a 1:1 stoichiometry.    If your standardised sodium hydroxide solution was determined to be 0.059 M, and it required an average titre (titration volume) of 28.1 mL, what is the concentration (in M) of the undiluted vinegar sample (the initial vinegar sample)?  Provide your answer to 2 significant figures. Flag question: Question 4Question 4Tips3 ptsOn domestic products such as vinegar, the concentration of active ingredients is rarely reported in Molar units.  It is more common to report the concentration as a composition such as mass per 100 mL or volume per 100 mL.  Vinegar is usually reported as a volume composition (volume of acetic acid per 100 mL vinegar - written as % v/v).  To convert a molar concentration to % v/v:Calculate how many moles of acetic acid is in 100 mL of solution (Hint: n = C x v)Calculate the mass of acetic acid from the amount of moles determined in step 1 (Hint: m= MW x n)Calculate the volume that the mass determined in step 2 would occupy.  To do this, you will have to divide the mass (determined in step 2) by the density of acetic acid (1.049 g.mL⁻¹).  The density states how much 1 mL of a liquid sample weighs - for acetic acid, 1 mL weighs 1.049 g.The answer in step 3 states what volume of acetic acid is present in a 100 mL sample of vinegar which equates to the composition we are after.  So we can just add "% v/v" to the end of our value to denote this is the volume of acetic acid per 100 mL of vinegar. If, from titration, you determined that concentration of acetic acid in vinegar is 0.504 M, what is the composition of acetic acid in vinegar (% v/v).  Answer to 3 significant figures (please enter numeric value only)