How many milliliters of a 0.266 m rbno3 solution are required to make 150.0 ml of 0.075 m rbno3 solution? how many milliliters of a 0.266 m rbno3 solution are required to make 150.0 ml of 0.075 m rbno3 solution? 42.3 ml 53.2 ml 35.1 ml 18.8 ml 23.6 ml

Answer : 42.3 ml of a 0.266 M solution are required.

Solution : Given,

Molarity of solution 1 = 0.266 M

Molarity of solution 2 = 0.075 M

Volume of solution 2 = 150 ml = 0.150 L      (1 L = 1000 ml)

Formula used :

where,

= Molarity of solution 1

= Molarity of solution 2

= Volume of solution 1  

= Volume of solution 2

Now put all the given values in above formula, we get

                (1 L = 1000 ml)

Therefore, 42.3 ml of a 0.266 M solution are required.

93.2 ml + 3920 ml = 4013 ml total  

(93.2 ml) * (2.03 m) / (4013 ml) = 0.0471 m

the volume in in l of the diluted solution would be 0.0471 m.

you need to take a look at the periodic system in order to sort these elements into pairs that would most likely exhibit similar chemical properties. first, you need to find elements that are found in the same group (groups are vertical in the periodic system), and then to see whether they have the same valence number. if they meet these criteria, they will have similar properties.

the first pair: ne (neon) and kr (krypton), are noble gases found in group 8

the second pair: li (lithium) and k (potassium), found in group 1

the third pair: s (sulfur) and te (tellurium), found in group 6