Which of the following NaCl solutions has the highest concentration (molarity) o

Question

Which of the following NaCl solutions has the highest concentration (molarity) of NaCl?

0.0175 mol NaCl in 1 L of solution

0.1 g NaCl in 100 mL of solution

0.5 g NaCl in 500 mL of solution

0.25 mol NaCl in 200 mL of solution

All have the same concentration

Which type of interaction must be strong in order to dissolve a salt?

Solvent-solvent

Solvent-solute

Solute-solute

Solvent-solution

All of the above

If the solubility of potassium nitrate at 20 °C is 37 g per 100 g of water, how many grams of the compound will dissolve in 1.0 mL of water?

37g

3.7g

0.37g

0.0037g

370g

Calculate the volume of 0.250 M solution made by dissolving 5.56 g of solid magnesium chloride in water.

0.234 mL

0.250 mL

1.39 mL

14.6 mL

234 mL

What is the mass (grams) of potassium nitrate (KNO3) in 500. mL of solution with a concentration of 4.90 M?

2.48 x 105g

991g

1.03 x 104g

248g

0.991g

Explanation / Answer

1) Molarity = moles per volume in litres

Hence, whichever solutions mol/V value is high, has the highest concentration of NaCl:

Various mol/V values of the question:

a) 0.0175 * /1 = 0.0175 M

b) As value of NaCl is given in grams, we convert it to moles first.

moles = given mass / molecular mass in grams

= 0.1 / 58.5 = 0.00171 moles

Hence, mol/V value = (0.00171 / 100ml ) = 1000 L / ml = 0.0171 M

c)0.5 g NaCl = 0.5 / 58.5 = 0.0085 moles

mol/V = 0.0085 / (500/ 1000 L) = 0.00425 M

d) mol/ V = 0.25 / (200 / 1000 L) = 1.25M

Obviously, as D has highest value of M, it is having the highest concentration.

2)Solvent - solute interaction must be greater than Solute-solute and Solvent-solvent interactions.

Hence, B is the answer.

3)Solubility of KNO3 = 37 g per 100 g of water

As density of water = 1 g / ml, Volume of 100 g water = 100 g / g * ml = 100 ml

Now, 37 g per 100 ml of water. Hence, 1 ml water dissolves : 1ml/100 ml * 37 g = 0.37 g of KNO3

Hence, C is the answer.

4)Number of moles in solid MgCl2 = mass / molecular mass = 5.56 / 95.2 (mol. wt of MgCl2 = 0.058 mol

So, 0.058 moles should contribute to a 0.25 M solution :

This happens when 0.058 moles are dissolved in volume V of the solution such that mol/V = 0.25

This implies: mol/V = 0.25

=> V = mol / 0.25 = 0.058/ 0.25 = 0.232 L = 232 ml of water

Hence, the closest answer is D

5) Moles of KNO3 Present in a 500ml 4.9M solution :

mol/ V = 4.9 M

moles = 4.9 * (500/1000 ) = 2.45 mol

mass of 1 mole KNO3 = 101 g / mol

mass of 2.45 mol  KNO3 = 101 * 2.45 g = 247.45 g

Hence, closest answer = C

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