Question

What is the molarity of a solution that contains 10.0 grams of NaOH in 1 L of solution?

How many grams of HCl are needed to make 500.0 mL of a 1.00 M solution (GMW HCl = 36.46 g)?

A 2 N HCl solution is also a ___ molar solution

Given a 1.5 M H₂SO₄ solution, what is the normality of this solution?

What is the gram equivalent weight of K₂CO₃ (K = 39.10 g, C = 12.01 g, O = 16.00 g

A total of 500 mL of a 55.0%^v/v ethanol (EtOH) solution is needed. In the stock room is a bottle of 95.0%^v/v EtOH. How much of the 95.0%^v/v EtOH should you use to prepare the 500-mL solution?

How many grams of HCl are needed to make a 1.5 molal solution in 1.0 kg of H₂O

Convert a 4 mg/dL Mg⁺² standard to mEq/L units (Mg = 24.30 g)

A 15.0%^w/v solution of CaCl₂ is needed. However, only CaCl₂ . 3H₂O is available. How many grams of CaCl₂ . 3H₂O are necessary to prepare the solution? (GMW CaCl₂ = 110.98)

How many milliliters of acetic acid (CH₃CO₂H) are needed to make a 5%^v/v solution using deionized water as the solvent

Answer 1

© To find or Molosity of solution which contain log of NaOH an 12 solution. Solution: Mass of NaOH = log Volume of wato = 1L Molas mass of NaOH = 40g/mol Mole of NooH = mass molas mass 10 Lo Imole of NaOH = 0.25 mol.) Molarity of NaOH = moles of solute volume of solent 0.25 Molarity of Noot = 0.25 M.

Given Molarity and volume of solution Solutions- IM hydrochloric aid solution contain Imol of HCL for one lehre. Since 500 mL is used then there will be ļ of the amount, 14 x 1 mol of Hel = 5oomt x - 103 HCL is present = 0.5 moles of in 500 ml solution. Converting to grams, = no of moles x molas mass = 0.58 36.46 L = 18.23 grums

3 A 2N HEL solution is also al solution. 2N is for an acid which releases aprotons from at. * N os nosmally soffered to apolon os hydroxide in . It is the cakon os anion seleased during hydrolysis.

- 6 Nosmalety = M x No of hydrogens (hydroxide Pons Given, 1.5M of H2SO4 N = 1.5% 2 hydrogen pons N = 3N. Normality of 1.5 M H₂SO4 solution is 30. ③ Gram equivalent weights - (Aromic mass x no of atoms) for each atom. Foә к, со = (39.10x)+(12.01 x) + (16 x 3) = 39.1 + 12.01 + 48 = 99.11 g/mol. 0 X 1