1.) Calculate the molarity of a solution prepared by dissolving 6.434 g of KCl in water to make a 1.500 L solution. 2.) If 10.0 g of silver nitrate is available, what volume of 0.25 M silver nitrate can be prepared?

3.) A laboratory procedure calls for a 0.250 M solution of cobalt (II) chloride. How many grams of cobalt (II) chloride are needed to make 0.500 L of this solution?

4.) 14.4 g of magnesium nitrate is dissolved in 1.20 L of water. 25 mL of this solution is taken and diluted to 175 mL with water. What is the final concentration of the nitrate ion?

(a)The molarity of KCl solution is, 0.9713 mole/L

(b)The molarity of  solution is, 0.00525 mole/L

(c)The molarity of  solution is, 0.0612 mole/L

(d)The molarity of  solution is, 7.61 mole/L

(e)The molarity of  solution is, 0.0565 mole/L

(f)The molarity of  solution is, 0.0113 mole/L

Explanation :

(a) 1.457 mol of KCl in 1.500 L of solution

Formula used :

Solute is KCl.

The molarity of KCl solution is, 0.9713 mole/L

(b) 0.515 gram of , in 1.00 L of solution

Formula used :

Solute is

Molar mass of  = 98 g/mole

The molarity of  solution is, 0.00525 mole/L

(c) 20.54 g of  in 1575 mL of solution

Formula used :

Solute is

Molar mass of  = 213 g/mole

The molarity of  solution is, 0.0612 mole/L

(d) 2.76 kg of  in 1.45 L of solution

Formula used :

Solute is

Molar mass of  = 250 g/mole

The molarity of  solution is, 7.61 mole/L

(e) 0.005653 mol of  in 10.00 ml of solution

Formula used :

Solute is .

The molarity of  solution is, 0.0565 mole/L

(f) 0.000889 g of glycine, , in 1.05 mL of solution

Formula used :

Solute is

Molar mass of  = 75 g/mole

The molarity of  solution is, 0.0113 mole/L

a polar bond is formed when electrons are shared unevenly between atoms