For an aqueous solution of glucose (C6H12O6), determine: A) The molarity of 2.00 L of a solution that contains 50.0 g of glucose

0.139 ML⁻¹

Explanation:

The Molarity (M) of a solution is the no of moles (amount) of a solute dissolved in a litre of solution.

M = no of moles / 1 litre of solution of solution

The first step is to calculate the no of moles of Glucose contained in 50g of it

No of moles of glucose  = mass/ molar mass of glucose

                                        = 50g / (180.156)g/mol

                                        = 0.27753≅0.278 moles of glucose

Therefore, the molarity of glucose in 2L of solution

                                         = 0.278 moles / 2 lites

                                         = 0.139M/Litre of glucose

                                         = 0.139 ML⁻¹ of glucose

fe 55.85 g/mol and h2o 18.02 g/mol.

the correct answer is b. a strong acid completely dissociates to its ions when dissolved in water. completely dissociates to and ions. a is incorrect because does not fully ionize to and ions and some remain as the compound . c and d are incorrect because, and do not fall into the category of bronsted lowry acids. a bronsted lowry acid is a proton donor and since both molecules do not have a proton to donate they are not bronsted lowry acids. likewise,   if we use the lewis acid definition, they cannot be acids. a lewis acid is an electron pair acceptor. both and have a complete octet so can not accept any electrons which disqualifies them as acids.