How many grams of solid NaOH are required to prepare 200-mL of a 0. 05 M solutio

Question

How many grams of solid NaOH are required to prepare 200-mL of a 0. 05 M solution? What will be the concentration from Problem 1 expressed in % w/v? How many milliliters of 5M NaCl are required to prepare 1500 mL of 0. 002 M NaCl? What will be the concentration of the diluted solution from Problem 3 expressed in millimolars, micromolars, and nanomolars? A solution contains 15 g of CaCl_2 in a total volume of 190 mL. Express the concentration in terms of grams/liter, % w/v, molars, and millimolars. Given stock solutions of glucose (1M), aparagine (100 mM) and NaH_2PO_4 (50 mM), how much of each solution do you need to prepare 500 mL of a reagent that contains 0.05 M glucose, 10 mM asparagine, and 2mM NaH_2PO_4? Calculate the number of millimoles in 500 mg of each of the following amino acids: alanine (MW = 89), leucine (MW = 131), tryptophan (MW = 204), cysteine (MW = 121), and glutamic acid (MW = 147). What molarity of HCl is needed so that 5mL diluted to 300 mL will yield 0.2 M? How much 0.2 M HCl can be made from 5.0 mL of 12.0 M HCl solution? What weight of glucose is required to prepare 2 L of a 5% w/v solution? How many milliliters of an 8.56% solution can be prepared from 42.8 g of sucrose? How many milliliters of CHCl_3 are needed to prepare a 2.5% v/v solution in 500 mL of methanol? If a 250-mL solution of ethanol in water is prepared with 4mL of absolute ethanol, what will be the concentration of ethanol in % v/v? For a list of web sites related to the material covered in this chapter, go to

Explanation / Answer

From the definition of molarity,

                                          Molarity(M) = (wt of the solute(g) * 1000)/ (Molecular weight of the solute * V)

                                     Given Molarity of the solution = 0.05 M

                                    Molecular weight of the Solute (NaOH) = 23+ 16 + 1 = 40 g/gmol

                                        Given Volume of water in which NaOH is diluted = 200 ml

                                         Let weight of the solute = w

                                      0.05 = (w*1000) / (40*200) ; Here 1000 is conversion factor from ml to L

                                           w = (0.05*40*200) /1000 = 0.4 g

So, 0.4 g of NaOH is required to prepare 200 mL of 0.05M NaOH solution

2) weight by volume percentage is one of the way of measuring or expressing the concentration

                     w/v % = weight of solute (g) * 100/ Volume of solution (mL)

                                    w/v %   = (0.4*100) / 200 = 0.2 w/v %

3) To calculate the volume of NaCl required to prepare 5 M Nacl: This is calculated using a known formula.

        Given : M1 = 0.002 M NaCl

                    V1 = 1500 mL

                  M2 = 5 M

                    V2 = ?

                       M1V1 = M2V2

                        0.002 * 1500 = 5 * V2

                          V2 = (0.002 * 1500) / 5 = 0.6 mL

       SO 0.6 mL of 5M Nacl solution is required to prepare 1500 mL of 0.002 M NaCl Solution

                                   

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