You make two dilutions. You take 83.52 mL of the stock solution and dilute to 500.0 mL to make Solution #1. You then add 52.27 mL of Solution #1 to 46.77 mL of water to make Solution #2. What is the concentration of the Solution #1 in terms of molarity?

The answer to the question is;

The concentration of the Solution #1 in terms of molarity is

0.16704X moles/litre.

Explanation:

Let the concentration of the stock solution be X moles/liter

Therefore, 83.52 ml of the stock solution contains

83.52×(X/1000) moles

Dilution of 83.52 ml of X to 500 ml gives solution 1 with a concentration of

500 ml of solution 1 contains 83.52×(X/1000) moles

Therefore 1000 ml or 1 litre contains 2×83.52×(X/1000) moles = 0.16704X moles/litre

The molarity of solution 1 is 0.16704X moles/litre.

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g